MARETONG: Kumpulan Soal UN / UNBK dan SBMPTN Logaritma serta Pembahasan

Tuesday, June 18, 2019

Kumpulan Soal UN / UNBK dan SBMPTN Logaritma serta Pembahasan

Soal dan Pembahasan Logaritma menjadi sesuatu yang sangat penting, mengingat persoalan logaritma ini menjadi sebuah persoalan yang sangat strategis karena selalu muncul dalam setiap Ujian Nasional maupun Ujian Masuk Perguruan Tinggi Negri. Soal Ujian Masuk PTN akan terasa hambar jika tidak ada soal logaritma. Mengingat dan mempertimbangkan hal tersebut, maka admin sengaja mengangkat topik ini dengan harapan adik-adik yang lagi berjuang untuk mendapatkan satu kursi di PTN idaman bisa tercapai. Soal-soal diambil dari soal-soal ujian nasional maupun Ujian Masuk PTN yang pernah diujikan. Supya tidak menjadi panjang lebar, silahkan pelajari soal dan pembahasan logaritma berikut.

Soal dan Pembahasan Logaritma
$1.$ Hasil dari $\displaystyle \dfrac{^3log36.^6log81\ +\ ^4log32}{^{\dfrac19}log27}$ adalah . . . .
$A.\ 11$
$B.\ 7$
$C.\ 4$
$D.\ -7$
$E.\ -11$
[Soal UNBK Matematika IPA 2018]
[Soal dan Pembahasan Logaritma]

$\displaystyle \dfrac{^3log36\ .\ ^6log81\ +\ ^4log32}{^{\dfrac19}log27}$
$= \displaystyle \dfrac{^3log6^2 .\ ^6log3^4\ +\ ^{\displaystyle 2^{2}}log2^5}{^{\displaystyle 3^{-2}}log3^3}$
$= \dfrac{2 . 4 .\ ^3log6\ .\ ^6log3\ +\ \dfrac52 .\ ^2log2}{\dfrac{3}{-2} .\ ^3log3}$
$= \dfrac{2.4 + \dfrac52}{-\dfrac{3}{2}}$
$= \dfrac{\dfrac{16 + 5}{2}}{-\dfrac{3}{2}}$
$= -\dfrac{21}{3}$
$= -7$
$Jawab:\ D.$

$Ingat\ !$
$^alogb\ .\ ^bloga = 1\ dan\ ^aloga = 1$

$2.$ Jika diketahui $^2log3 = x$, maka nilai dari $^8log12$ adalah . . . .
$A.\ \dfrac{-x - 2}{3}$
$B.\ \dfrac{x - 2}{3}$
$C.\ \dfrac{x + 2}{3}$
$A.\ \dfrac{x + 3}{2}$
$A.\ \dfrac{x - 3}{2}$
[Soal UNBK Matematika IPS 2018]
[Soal dan Pembahasan Logaritma]

$^8log12 =\ ^{\displaystyle 2^{3}log3.4}$
$=\ ^{\displaystyle 2^{3}}log3 +\ ^{\displaystyle 2^{3}}log4$
$=\ ^{\displaystyle 2^{3}}log3 +\ ^{\displaystyle 2^{3}}log2^2$
$= \dfrac{1}{3}\ .\ ^2log3 + \dfrac{2}{3}\ .\ ^2log2$
$= \dfrac13x + \dfrac23$
$= \dfrac{x + 2}{3}$
$Jawab:\ C$

$3.$ Hasil $\dfrac{^\sqrt{3}log5\ .\ ^{25}log3\sqrt{3}\ -\ ^4log16}{^3log54\ -\ ^3log2}$ adalah . . . .
$A.\ -\dfrac92$
$B.\ -\dfrac16$
$C.\ -\dfrac13$
$D.\ 3$
$E.\ \dfrac92$
[Soal UNBK Matematika IPA 2017]
[Soal dan Pembahasan Logaritma]

$\dfrac{^\sqrt{3}log5\ .\ ^{25}log3\sqrt{3}\ -\ ^4log16}{^3log54\ -\ ^3log2}$
$= \dfrac{^{\displaystyle 3^{\frac12}}log5\ .\ ^{\displaystyle 5^{2}}log3^{\frac{3}{2}}\ -\ ^{\displaystyle 2^{2}}log2^4}{^3log\dfrac{54}{2}}$
$= \dfrac{\dfrac32.\dfrac12.\dfrac{1}{\frac12}\ .\ ^3log5\ .\ ^5log3\ -\ 4.\dfrac12\ .\ ^2log2}{^3log3^3}$
$= \dfrac{\dfrac32\ .\ ^3log5\ .\ ^5log3\ -\ 2\ .\ ^2log2}{3\ .\ ^3log3}$
$= \dfrac{\dfrac32 - 2}{3}$
$= \dfrac{-\dfrac{1}{2}}{3}$
$= -\dfrac16$
$Jawab:\ B.$

$4.$ Nilai dari $^7log4\ .\ ^2log5\ +\ ^7log\dfrac{49}{25} =$ . . . .
$A.\ 1$
$B.\ 2$
$C.\ 3$
$D.\ 4$
$E.\ 5$
[Soal UNBK Matematika IPS 2017]
[Soal dan Pembahasan Logaritma]

$^7log4\ .\ ^2log5\ +\ ^7log\dfrac{49}{25}$
$=\ ^7log2^2\ .\ ^2log5\ +\ ^7log49\ -\ ^7log25$
$= 2\ .\ ^7log2\ .\ ^2log5 +\ ^7log7^2\ -\ ^7log5^2$
$= 2.\dfrac{log2}{log7}\ .\ \dfrac{log5}{log2}\ +\ 2.^7log7 - 2.^7log5$
$= 2.\dfrac{log5}{log7} + 2.1 - 2.^7log5$
$= 2.^7log5 + 2 - 2.^7log5$
$= 2$
$Jawab:\ B.$

$5.$ Nilai dari $\left(\dfrac{^5log\ 9\ .\ ^{81}log\ 625 + ^5log\ 125}{^6log\ 216 - ^6log\ 36} \right)^2 =$. . . .
$A.\ 625$
$B.\ 125$
$C.\ 25$
$D.\ -25$
$E.\ -125$
[Soal UNBK Matematika IPA 2016]
[Soal dan Pembahasan Logaritma]

$\left(\dfrac{^5log\ 9\ .\ ^{81}log\ 625 +\ ^5log\ 125}{^6log\ 216\ -\ ^6log\ 36} \right)^2$
$= \left(\dfrac{^5log\ 3^2\ .\ ^{\displaystyle 3^4}log\ 5^4 +\ ^5log\ 5^3}{^6log\ \dfrac{216}{36}} \right)^2$
$= \left(\dfrac{\dfrac44.2.^5log\ 3\ .\ ^3log\ 5 +\ 3.^5log\ 5}{^6log\ 6} \right)^2$
$= \left(\dfrac{2 + 3}{1}\right)^2$
$= 5^2$
$= 25$
$Jawab:\ C.$

$6.$ Nilai $x$ yang memenuhi $^{\frac{1}{3}}log(x + \sqrt{3})\ +\ ^{\frac{1}{3}}log(x - \sqrt{3}) > 0$ adalah. . . .
$A.\ x < -\sqrt{3}\ atau\ 0 < x < 2$
$B.\ -2 < x < -\sqrt{3}\ atau\ \sqrt{3} < x < 2$
$C.\ \sqrt{3} < x < 2$
$D.\ -2 < x < 2$
$E.\ -\sqrt{3} < x < 2$
[Soal UNBK Matematika IPA 2016]
[Soal dan Pembahasan Logaritma]

$Syarat\ pertidaksamaan:$
Apabila $a$ adalah basis dan $0 < a < 1,$
jika $^alog\ f(x) >\ ^alog\ g(x)$ maka $f(x) < g(x)$
$^{\frac{1}{3}}log(x + \sqrt{3})\ +\ ^{\frac{1}{3}}log(x - \sqrt{3}) > 0$
$^{\frac{1}{3}}log(x + \sqrt{3})(x - \sqrt{3}) > 0$
$^{\frac{1}{3}}log(x + \sqrt{3})(x - \sqrt{3}) >\ ^{\frac{1}{3}}log\ 1$
$(x + \sqrt{3})(x - \sqrt{3}) < 1$
$x^2 - 3 < 1$
$x^2 - 4 < 0$
$(x + 2)(x - 2) < 0$
$-2 < x < 2$ . . . . *

$Syarat\ numerus:$
$x + \sqrt{3} > 0$
$x > -\sqrt{3}$ . . . . **

$x - \sqrt{3} > 0$
$x > \sqrt{3}$ . . . . ***


$*\ \cap\ **\ \cap\ ***\ → \sqrt{3} < x < 2$
$Jawab:\ C.$

$7.$ Nilai $3(^2log\ y) -\ ^2log\ y^2 + 2log\dfrac 1y$ adalah . . . .
$A.\ 1$
$B.\ 0$
$C.\ y$
$D.\ -1$
$E.\ -y$
[Soal UNBK Matematika IPS 2016]
[Soal dan Pembahasan Logaritma]

$3(^2log\ y) -\ ^2log\ y^2 +\ ^2log\dfrac1y$
$ = (^2log\ y^3) -\ ^2log\ y^2 +\ ^2log\ y^{-1}$
$=\ ^2log\dfrac{y^3.y^{-1}}{y^2}$
$=\ ^2log\ y^{3-1-2}$
$=\ ^2log\ y^0$
$=\ ^2log\ 1$
$= 0$
$Ingat\ !$
$log\ 1 = 0$
$Jawab:\ B.$

$8.$ Diketahui $^plog2 = 3$ dan $^qlog3 = 2.$ Jika $x = p^3$ $y = q^2$, maka $^xlog\ y =$ . . . .
$A.\ \dfrac{log\ 3}{log\ 2}$
$B.\ \dfrac{3log\ 3}{2log\ 2}$
$C.\ \dfrac{9log\ 3}{4log\ 2}$
$D.\ \dfrac{2log\ 2}{3log\ 3}$
$E.\ \dfrac{log\ 2}{log\ 3}$
[Soal SBMPTN 2015 Matdas]
[Soal dan Pembahasan Logaritma]

$p^3 = 2$
$x = p^3$
$x = 2$

$q^2 = 3$
$y = q^2$
$y = 3$

$^xlog\ y =\ ^2log3 = \dfrac{log\ 3}{log\ 2}$
$Jawab:\ A.$

$9.$ Diketahui $^alog2 = x$, dan $^alog5 = y$, maka $log\ a^{2x} + 2log\ a^y =$ . . . .
$A.\ 1$
$B.\ 2$
$C.\ 3$
$D.\ 4$
$E.\ 5$
[Soal SBMPTN 2015 Matdas]
[Soal dan Pembahasan Logaritma]

$^alog\ 2 = x → 2 = a^x$
$^alog\ 5 = y → 5 = a^y$
$log\ a^{2x} + 2log\ a^y = log\ (a^x)^2 + log\ (a^y)^2$
$= log\ 2^2 + log\ 5^2$
$= log\ 4.25$
$= log\ 100$
$= 2$
$Jawab:\ B.$

$10.$ Diketahui $a =\ ^4log\ x$ dan $b =\ ^2log\ x$. Jika $^4log\ b +\ ^2log\ a = 2$, maka $a + b$ adalah . . . .
$A.\ 4$
$B.\ 6$
$C.\ 8$
$D.\ 12$
$E.\ 16$
[Soal SBMPTN 2014 Matdas]
[Soal dan Pembahasan Logaritma]

$a =\ ^4log\ x $
$a =\ ^{\displaystyle 2^2}log\ x$
$a = \dfrac12.\ ^2log\ x$ . . . . *
$b =\ ^2log\ x$ . . . . **
$Dari\ pers\ *\ dan\ **$
$a = \dfrac12b$ . . . . ***
$^4log\ b +\ ^2log\ a = 2$
$^{\displaystyle 2^2}log\ b +\ ^2log\ a = 2$
$\frac12.\ ^2log\ b +\ ^2log\ a = 2$
$^2log\ b^{\frac12} +\ ^2log\ a = 2$
$^2log\ b^{\frac12}.a = 2$
$b^{\frac12}.a = 2^2$
$b^{\frac12}.a = 4$ . . . . ****
$Dari\ pers\ ***\ dan\ ****$
$\displaystyle b^{\frac12}.\dfrac12b = 4$
$b^{\frac32} = 8$
$b = 8^{\frac23}$
$b = \sqrt[3]{8^2}$
$b = \sqrt[3]{64}$
$b = 4$

$a = \dfrac12b$
$= \dfrac12.4$
$= 2$

$a + b = 2 + 4 = 6$
$Jawab:\ B.$

$11.$ Jika $^blog\ a = -2$ dan $^3log\ b = (^3log\ 2)(1 +\ ^2log\ 4a),$ maka $4a + b =$ . . . .
$A.\ 768$
$B.\ 72$
$C.\ 36$
$D.\ 12$
$E.\ 3$
[Soal SBMPTN 2014 Matdas]
[Soal dan Pembahasan Logaritma]

$^3log\ b = (^3log\ 2)(1 +\ ^2log\ 4a)$
$\dfrac{log\ b}{log\ 3} = \dfrac{log\ 2}{log\ 3}(1 +\ ^2log\ 4a)$
$[log\ 3\ diruas\ kiri\ dan\ kanan\ bisa\ dicoret]$
$\dfrac{log\ b}{log\ 2} = (^2log\ 2 +\ ^2log\ 4a)$
$^2log\ b =\ ^2log\ 2.4a$
$b = 8a$ . . . . *

$^blog\ a = -2 → a = b^{-2}$
$a = (8a)^{-2}$
$a = \dfrac{1}{(8a)^2}$
$a = \dfrac{1}{64a^2}$
$a.a^2 = \dfrac{1}{64}$
$a = \sqrt[3]{\dfrac{1}{64}}$
$a = \dfrac14$

$b = 8a = 8.\dfrac14 = 2$

$4a + b = 4.\dfrac14 + 2 = 3$
$Jawab:\ E.$

$12.$ Jika $\dfrac{^3log\ x}{^3log\ w} = 2$, dan $^{xy}log\ w = \dfrac25,$ maka nilai $\dfrac{^2log\ w}{^2log\ y}$ adalah . . . .
$A.\ 8$
$B.\ 6$
$C.\ 4$
$D.\ 2$
$E.\ 1$
[Soal SBMPTN 2013 Matdas]
[Soal dan Pembahasan Logaritma]

$\dfrac{^3log\ x}{^3log\ w} = 2 →\ ^wlog\ x = 2$ . . . . *
$Ingat:\ \dfrac{^alog\ p}{^alog\ q} =\ ^qlog\ p$

$^{xy}log\ w = \dfrac25$
$\dfrac{log\ w}{log\ x + log\ y} = \dfrac25$
$Tambahkan\ bilangan\ pokok\ w\ !$
$\dfrac{^wlog\ w}{^wlog\ x + ^wlog\ y} = \dfrac25$ . . . . **
Dari pers * dan **
$\dfrac{1}{2 +\ ^wlog\ y} = \dfrac25$
$5 = 4 +\ 2.^wlog\ y$
$5 - 4 =\ 2.^wlog\ y$
$\dfrac12 =\ ^wlog\ y$

$Ditanya:\ \dfrac{^2log\ w}{^2log\ y} =\ ^ylog\ w =\ ?$
$^ylog\ w = \dfrac{1}{^wlog\ y} = \dfrac{1}{\dfrac12} = 2$
$Jawab:\ D.$

$13.$ Jika $2\left(^xlog\dfrac{1}{3^x + 2} \right)\left(^3log\ \dfrac1x \right) = 2 + x$ maka $(27)^x =$ . . . .
$A.\ 8$
$B.\ 9$
$C.\ 27$
$D.\ 64$
$E.\ 125$
[UM UGM 2018 Matdas]
[Soal dan Pembahasan Logaritma]

$2\left(^xlog\dfrac{1}{3^x + 2} \right)\left(^3log\ \dfrac1x \right) = 2 + x$
$2\left(^xlog(3^x + 2)^{-1} \right)\left(^3log\ x^{-1} \right) = 2 + x$
$2.(-1).(-1).\left(^xlog(3^x + 2) \right)\left(^3log\ x \right) = 2 + x$
$2.\left(^xlog(3^x + 2) \right)\left(^3log\ x \right) = 2 + x$
$\left(^xlog(3^x + 2) \right)\left(^3log\ x \right) = 1 + \dfrac{x}{2}$
$Ingat:\ ^alog\ b\ .\ ^clog\ a =\ ^clog\ b$
$^3log\ (3^x + 2) = 1 + \dfrac{x}{2}$
$3^x + 2 = 3^{1 + \frac{x}{2}}$
$3^x + 2 = 3.3^{\frac{x}{2}}$
$\left(3^{\frac{x}{2}} \right)^2 + 2 = 3.3^{\frac{x}{2}}$
$\left(3^{\frac{x}{2}} \right)^2 - 3.3^{\frac{x}{2}} + 2 = 0$
$(3^{\frac{x}{2}} - 2)(3^{\frac{x}{2}} - 1) = 0$
$3^{\frac{x}{2}} = 2\ atau\ 3^{\frac{x}{2}} = 1$
$Ditanya:\ (27)^x = ?$
$(27)^x = \left(3^{\frac{x}{2}}\right)^6$
$= 2^6\ atau\ 1^6$
$= 64\ atau\ 1$
$Jawab:\ D.$

$14.$ Jika $a$ dan $b$ memenuhi sistem persamaan $\begin{cases} \dfrac{3}{log\ a} + \dfrac{4}{log\ b} = 7 \\ -\dfrac{1}{log\ a} + \dfrac{2}{log\ b} = 11 \end{cases}$ $Maka\ ^alog\ \dfrac{1}{b} + ^blog\ \dfrac{1}{a} =$ . . . .
$A.\ \dfrac16$
$B.\ \dfrac{7}{12}$
$C.\ 1\dfrac16$
$D.\ 2\dfrac{1}{12}$
$E.\ 2\dfrac14$
[Soal UM UGM 2018 Matdas]
[Soal dan Pembahasan Logaritma]

$\dfrac{3}{log\ a} + \dfrac{4}{log\ b} = 7$ . . . . *
$-\dfrac{1}{log\ a} + \dfrac{2}{log\ b} = 11$ . . . . **

Eliminasi pers * dan pers ** dengan mengalikan
pers ** dengan $3$ terlebih dahulu.

$\dfrac{3}{log\ a} + \dfrac{4}{log\ b} = 7$
$-\dfrac{3}{log\ a} + \dfrac{6}{log\ b} = 33$
---------------------------------------------- +
$\dfrac{10}{log\ b} = 40 → log\ b = \dfrac14$ . . . . ***
Substitusikan $log\ b = \dfrac14$ ke dalam pers *
$\dfrac{3}{log\ a} + \dfrac{4}{\dfrac14} = 7$
$\dfrac{3}{log\ a} + 16 = 7$
$\dfrac{3}{log\ a} = -9 → log\ a = -\dfrac13$

$Ditanya:\ ^alog\ \dfrac{1}{b} + ^blog\ \dfrac{1}{a} =\ ?$
$^alog\ \dfrac{1}{b} + ^blog\ \dfrac{1}{a} = -^alog\ b + (-^blog\ a)$
$= -\dfrac{log\ b}{log\ a} - \dfrac{log\ a}{log\ b}$
$= -\dfrac{\dfrac14}{-\dfrac13} - \dfrac{-\dfrac13}{\dfrac14}$
$= \dfrac34 + \dfrac43$
$= \dfrac{25}{12}$
$= 2\dfrac14$
$Jawab:\ D.$

$15.$ Jika $2^4log\ x -\ ^4log\ (4x + 3) = -1$, maka $^2log\ x =$ . . . .
$A.\ ^2log\ 3 - 1$
$B.\ ^2log\ 3 + 1$
$C.\ 1 -\ ^2log\ 3$
$D.\ -1 -\ ^2log\ 3$
$A.\ ^2log\ 3 +\ ^3log\ 2$
[Soal UM UGM 2018 Matematika IPA]
[Soal dan Pembahasan Logaritma]

$2^4log\ x -\ ^4log\ (4x + 3) = -1$
$^4log\ x^2 -\ ^4log\ (4x + 3) = -1$
$^4log\ \dfrac{x^2}{4x + 3} =\ ^4log\ \dfrac14$
$\dfrac{x^2}{4x + 3} = \dfrac14$
$4x^2 = 4x + 3$
$4x^2 - 4x - 3 = 0$
$(2x + 1)(2x - 3) = 0$
$x = -\dfrac12\ atau\ \dfrac32$

Karena yang ditanya adalah $^2log\ x$, maka $x > 0$ $(Syarat\ numerus)$ Berarti yang memenuhi adalah $x = \dfrac32$
$^2log\ x =\ ^2log\ \dfrac32 =\ ^2log\ 3 -\ ^2log\ 2$
$=\ ^2log\ 3 - 1$
$Jawab:\ A.$

$16.$ Pertidaksamaan $^2log\ (x^2 - x) \leq 1$, mempunyai penyelesaian . . . .
$A.\ x < 0\ atau\ x > 1$
$B.\ -1 < x < 2;\ x \ne 1;\ x \ne 0$
$C.\ -1 \leq x < 0\ atau\ 1 < x \leq 2$
$D.\ -1 \leq x \leq 0\ atau\ 1 \leq x \leq 2$
$E.\ -1 < x < 0\ atau\ 1 \leq x < 2$
[Soal UM UGM 2018 Matematika IPA]
[Soal dan Pembahasan Logaritma]

$^2log\ (x^2 - x) \leq 1$
$^2log\ (x^2 - x) \leq\ ^2log\ 2$

$Syarat\ pertidaksamaan$
$x^2 - x \leq 2$
$x^2 - x - 2 \leq 0$
$(x + 1)(x - 2) \leq 0$
$-1 \leq x \leq 2$ . . . . *

$Syarat\ numerus$
$x^2 - x > 0$
$x(x - 1) > 0$
$x < 0\ atau\ x > 1$ . . . . **


$*\ \cap\ **\ →\ -1 \leq x < 0\ atau\ 1 < x \leq 2$
$Jawab:\ C.$

$17.$ Jika $x > y ≥ 1$ dan $log(x^2 + y^2 + 2xy) = 2log(x^2 - y^2)$ maka $^xlog\ (1 + y) =$ . . . .
$A.\ log\ 2$
$B.\ -1$
$C.\ -\dfrac12$
$D.\ \dfrac12$
$E.\ 1$
[Soal UM UGM 2018 Matematika IPA]
[Soal dan Pembahasan Logaritma]

$log(x^2 + y^2 + 2xy) = 2log(x^2 - y^2)$
$log(x^2 + y^2 + 2xy) = log(x^2 - y^2)^2$
$(x^2 + y^2 + 2xy) = (x^2 - y^2)^2$
$(x + y)^2 = (x^2 - y^2)(x^2 - y^2)$
$(x + y)^2 = (x + y)(x - y)(x + y)(x - y)$
$(x + y)^2 = (x + y)^2(x - y)^2$
$1 = (x - y)^2$
$x - y = \pm 1$
$Karena\ x > y \geq 1,\ maka:$
$x - y = 1 → y = x - 1$ . . . . *
$^xlog\ (1 + y) =\ ^xlog\ (1 + x - 1)$
$= ^xlog\ x = 1$
$Jawab:\ E.$

$18.$ Jika $^3log\ x +\ ^4log\ y^2 = 5$, maka nilai maksimum dari $^3log\ x\ .\ ^2log\ y$ adalah . . . .
$A.\ \dfrac{25}{4}$
$B.\ \dfrac{25}{9}$
$C.\ \dfrac{25}{16}$
$D.\ 1$
$E.\ \dfrac{25}{36}$
[Soal UM UGM 2017 Matematika IPA]
[Soal dan Pembahasan Logaritma]

$^3log\ x +\ ^4log\ y^2 = 5$
$^3log\ x = 5 -\ ^4log\ y^2$
$^3log\ x = 5 - \dfrac22\ .\ ^2log\ y$
$^3log\ x = 5 -\ ^2log\ y$

$^3log\ x\ .\ ^2log\ y = (5 -\ ^2log\ y)^2log\ y$
$Misalkan:\ ^2log\ y = p$
$Misalkan:\ ^3log\ x\ .\ ^2log\ y = f(p)$
$f(p) = (5 - p)p$
$f(p) = 5p - p^2$
Karena $p = -1 < 0$, maka $f(p) = 5p - p^2$ adalah fungsi kuadrat terbuka ke bawah.
$Nilai\ maksimum = f\left(\dfrac{-b}{2a}\right)$
$= f\left(\dfrac{-5}{2.(-1)}\right)$
$= f\left(\dfrac52\right)$
$= 5.\dfrac52 - \left(\dfrac52 \right)^2$
$= \dfrac{25}{2} - \dfrac{25}{4}$
$= \dfrac{50}{4} - \dfrac{25}{4}$
$= \dfrac{25}{4}$
$Jawab:\ A.$

$19.$ Jika $^7log(^3log(^2log\ x)) = 0$, Nilai $2x +\ ^4log\ x^2$ adalah . . . .
$A.\ 10$
$B.\ 12$
$C.\ 19$
$D.\ 21$
$E.\ 24$
[Soal SIMAK UI 2018 Matdas]
[Soal dan Pembahasan Logaritma]

$^7log(^3log(^2log\ x)) = 0$
$^7log(^3log(^2log\ x)) =\ ^7log\ 1$
$^3log(^2log\ x) = 1$
$^3log(^2log\ x) =\ ^3log\ 3$
$^2log\ x = 3$
$x = 2^3 = 8$

$2x +\ ^4log\ x^2 = 2.8 +\ ^4log\ 8^2$
$= 16 +\ ^4log\ 64$
$= 16 +\ ^4log\ 4^3$
$= 16 + 3\ .\ ^4log\ 4$
$= 19$
$Jawab:\ C.$

$20.$ Himpunan penyelesaian dari
$^xlog\ 2 + 1 \geq\ ^xlog(x^2 - 2x + 4)$ adalah . . . .
$A.\ ∅$
$B.\ \text{{x | x > 0}}$
$C.\ \text{{x | 0 < x < 1}}$
$D.\ \text{{x | 0 < x < 1}}\ atau\ x \ne 2$
$E.\ \text{{x | 0 < x < 1}}\ atau\ x = 2$
[Soal SIMAK UI 2010 Matematika IPA]
[Soal dan Pembahasan Logaritma]

$^xlog\ 2 + 1 \geq\ ^xlog(x^2 - 2x + 4)$
$^xlog\ 2 +\ ^xlog\ x \geq\ ^xlog(x^2 - 2x + 4)$
$^xlog\ 2x \geq\ ^xlog(x^2 - 2x + 4)$

$Syarat\ numerus$
$(i).\ x > 0$
$(ii).\ x^2 - 4x + 4 > 0$
$(x - 2)^2 > 0$ ← selalu benar untuk $x \ne 2$
$Dari\ (i)\ dan\ (ii)$
$x > 0\ dan\ x \ne 2$ . . . . *.

$Syarat\ pertidaksamaan$
$Solusi\ I$
$(i).$ Jika $0 < x < 1$, pertidaksamaan menjadi:

$(ii).\ 2x \leq x^2 - 2x + 4$
$0 \leq x^2 - 4x + 4$
$x^2 - 4x + 4 \geq 0$
$(x - 2)^2 \geq 0$ ← selalu benar untuk semua nilai $x$
$(i)\ \cap\ (ii) = 0 < x < 1$ . . . . **.

$Solusi\ II$
$(i).$ Jika $x > 1$, pertidaksamaan menjadi:
$(ii).\ 2x \geq x^2 - 2x + 4$
$0 \geq x^2 - 4x + 4$
$x^2 - 4x + 4 \leq 0$
$(x - 2)^2 \leq 0$ ← hanya benar jika $x = 2$
$(i)\ \cap\ (ii) → x = 2$ . . . . ***

$*\ \cap\ (**\ \cup\ ***) → 0 < x < 1$
$Jawab:\ C.$

$21.$ Himpunan penyelesaian pertidaksamaan $log|x + 1| \geq log\ 3 + log|2x - 1|$ adalah . . . .
$A.\ \begin{Bmatrix} x ∈ R\ |\ \dfrac27 \leq x \leq \dfrac45,\ x \ne \dfrac12 \end{Bmatrix}$
$B.\ \begin{Bmatrix} x ∈ R\ |\ \dfrac12 \leq x \leq \dfrac45\end{Bmatrix}$
$C.\ \begin{Bmatrix} x ∈ R\ |\ \dfrac27 \leq x \leq \dfrac45\end{Bmatrix}$
$D.\ \begin{Bmatrix} x ∈ R\ |\ x \leq 1\ atau\ x > \dfrac12\end{Bmatrix}$
$E.\ \begin{Bmatrix} x ∈ R\ |\ x \leq \dfrac45, x \ne \dfrac12\end{Bmatrix}$
[Soal SIMAK UI 2014 Matematika IPA]
[Soal dan Pembahasan Logaritma]

$Syarat\ numerus$
$1.\ | x + 1| > 0 → x \ne -1$ . . . . *
(Selalu terpenuhi untuk semua nilai $x$, kecuali $x = -1$)
$2.\ |2x - 1| > 0 → x \ne \dfrac12$ . . . . **
(Selalu terpenuhi untuk semua nilai $x$, kecuali $x = \dfrac12)$

$Syarat\ pertidaksamaan$
$log|x + 1| \geq log\ 3 + log|2x - 1|$
$log|x + 1| \geq log|2x - 1|3$
$|x + 1| \geq |6x - 3|$
$(x + 1)^2 \geq (6x - 3)^2$
$(x + 1)^2 - (6x - 3)^2 \geq 0$
$[(x + 1) + (6x - 3)][(x + 1) - (6x - 3)] \geq 0$
$(7x - 2)(-5x + 4) \geq 0$
$(7x - 2)(5x - 4) \leq 0$
$\dfrac27 \leq x \leq \dfrac45$ . . . . ***

$*\ \cap\ **\ \cap\ ***$
$\dfrac27 \leq x \leq \dfrac45,\ x \ne \dfrac12$
$Jawab:\ A.$

$22.$ Jika $2^a = 3$, $3^b = 4$, $4^c = 5$, $5^d = 6$,
$6^e = 7$, $7^f = 8$, maka $abcdef =$ . . . .
$A.\ 2$
$B.\ 3$
$C.\ 4$
$D.\ 8$
$E.\ 16$
[Soal SIMAK UI 2017 Matdas]
[Soal dan Pembahasan Logaritma]

$^2log\ 3 = a$, $^3log\ 4 = b$, $^4log\ 5 = c$, $^5log\ 6 = d$
$^6log\ 7 = e$, $^7log\ 8 = f.$
$^2log3.^3log4.^4log5.^5log6.^6log7.^7log8 = abcdef$
$^2log8 = abcdef$
$^2log2^3 = abcdef$
$3.^2log2 = abcdef$
$3 = abcdef$
$Jawab:\ B.$

$23.$ Jika $x$ dan $y$ memenuhi $logx^3 - logy^2 = 4$, dan $logx^4 + logy^3 = 11$, maka $y^2 - x =$ . . . .
$A.\ 0$
$B.\ 10$
$C.\ 900$
$D.\ 1900$
$E.\ 8000$
[Soal SIMAK UI 2017 Matdas]
[Soal dan Pembahasan Logaritma]

$logx^3 - logy^2 = 4$
$3logx - 2logy = 4$ . . . . *

$logx^4 + logy^3 = 11$
$4logx + 3logy = 11$ . . . . **

Eliminasi persamaan * dan persamaan **
$9logx - 6logy = 12$
$8logx + 6logy = 22$
--------------------- +
$17logx = 34$
$logx = 2$ . . . . ***
$x = 10^2$
$x = 100$

Substitusikan $logx = 2$ ke pers **
$4.2 + 3logy = 11$
$3logy = 3$
$logy = 1$
$y = 10^1$
$y = 10$

$y^2 - x = 10^2 - 100$
$= 100 - 100 = 0$
$Jawab:\ A.$

$24.$ Diketahui $a$ dan $b$ adalah bilangan bulat positif yang tidak sama dengan satu dan persamaan $^alogx.^blogx = \dfrac{^xlogb}{^xloga}$. Nilai $(a + b)x$ adalah . . . .
$A.\ ab + b^2\ atau\ \dfrac{a}{b} + 1$
$B.\ a^2b + ab\ atau\ \dfrac{a^2}{b} + a$
$C.\ ab + a^2 \ atau\ \dfrac{b}{a} + 1$
$D.\ ab + ab^2\ atau\ \dfrac{b^2}{a} + a$
$E.\ 2a + 2b^2\ atau\ \dfrac{a}{2} + \dfrac{b}{2}$
[Soal SIMAK UI 2015 Matdas]
[Soal dan Pembahasan Logaritma]

$^alogx.^blogx = \dfrac{^xlogb}{^xloga}$
$\dfrac{logx}{loga}.\dfrac{logx}{logb} = \dfrac{logb}{loga}$
$\dfrac{log^2x}{loga.logb} = \dfrac{logb}{loga}$
$log^2x = \dfrac{logb}{loga}loga.logb$
$log^2x = log^2b$
$\left(\dfrac{logx}{logb}\right)^2 = 1$
$\dfrac{logx}{logb} = \pm 1$
$^blogx = \pm 1$
$x_1 = b^1 = b$
$x_2 = b^{-1} = \dfrac{1}{b}$

$(a + b)x = (a + b)b$ atau $(a + b)\dfrac{1}{b}$
$= ab + b^2$ atau $\dfrac{a}{b} + 1$
$Jawab:\ A.$

$25.$ Jika $^{ab}loga = 4$, maka $^{ab}log\dfrac{\sqrt[3]{a}}{\sqrt{b}} =$ . . . .
$A.\ -3$
$B.\ -\dfrac34$
$C.\ -\dfrac16$
$D.\ \dfrac{29}{42}$
$E.\ \dfrac{17}{6}$
[Soal SIMAK UI 2014 Matdas]
[Soal dan Pembahasan Logaritma]

$^{ab}loga = 4 → \dfrac{loga}{logab} = 4$
$\dfrac{loga}{loga + logb} = 4$
$Ambil\ bilangan\ pokok\ a\ !$
$\dfrac{^aloga}{^aloga + ^alogb} = 4$
$\dfrac{1}{1 + ^alogb} = 4$
$1 = 4 + 4^alogb$
$^alogb = -\dfrac34$ . . . . *

$^{ab}log\dfrac{\sqrt[3]{a}}{\sqrt{b}} =\ ^{ab}loga^{\frac13} -\ ^{ab}logb^{\frac12}$
$= \dfrac{\dfrac13loga}{logab} - \dfrac{\dfrac12logb}{logab}$
$= \dfrac{\dfrac13loga}{loga + logb} - \dfrac{\dfrac12logb}{loga + logb}$
$Ambil\ bilangan\ pokok = a\ !$
$= \dfrac{\dfrac13.^aloga}{^aloga + ^alogb} - \dfrac{\dfrac12.^alogb}{^aloga + ^alogb}$
$= \dfrac{\dfrac13.1}{1 - \dfrac34} - \dfrac{\dfrac12.(-\dfrac34)}{1 - \dfrac34}$
$= \dfrac{\dfrac13}{\dfrac14} - \dfrac{-\dfrac{3}{8}}{\dfrac14}$
$= \dfrac43 + \dfrac32$
$= \dfrac{17}{6}$
$Jawab:\ E.$

$26.$ Diketahui bahwa $^3logx.^6logx.^9logx =$ $^3logx.^6logx + ^3logx.^9logx + ^6logx.^9logx$, maka Nilai $x$ adalah . . . .
$1.\ \dfrac13$
$2.\ 1$
$3.\ 48$
$4.\ 162$
[Soal SIMAK UI 2013 Matdas]
[Soal dan Pembahasan Logaritma]

$\dfrac{logx.logx.logx}{log3.log6.log9} = \dfrac{logx.logx}{log3.log6} + \dfrac{logx.logx}{log3.log9} + \dfrac{logx.logx}{log6.log9}$
$= log^2x\left(\dfrac{1}{log3.log6} + \dfrac{1}{log3.log9} + \dfrac{1}{log6.log9}\right)$
$= log^2x\left(\dfrac{log9}{log3.log6.log9} + \dfrac{log6}{log3.log6.log9} + \dfrac{log3}{log3.log6.log9}\right)$
$\dfrac{log^3x}{log3.log6.log9} = \dfrac{log^2x}{log3.log6.log9}\left(log9 + log6 + log3\right)$
$log^3x = log^2x\left(log9 + log6 + log3\right)$
$log^3x - log^2x\left(log9 + log6 + log3\right) = 0$
$log^2x(logx - \left(log9 + log6 + log3\right)) = 0$
$log^2x = 0\ atau\ log\ x = log9 + log6 + log3$
$log\ x = 0\ atau\ log\ x = log\ 9.6.3$
$log\ x = log\ 1\ atau\ log\ x = log\ 162$
$x = 1\ atau\ x = 162$
$Opsi\ yang\ benar\ adalah\ 2\ dan\ 4$
$Jawab: C.$

$27.$ Nilai $x$ yang memenuhi $2logx \leq log(3x + 7) + 2log\ 2$ adalah . . . .
$A.\ -2 \leq x \leq 14$
$B.\ -2 \leq x \leq 0$
$C.\ 0 < x \leq 14$
$D.\ -2 < x < 0$
$E.\ 0 \leq x \leq 14$
[Soal SIMAK UI 2012 Matdas]
[Soal dan Pembahasan Logaritma]

$Syarat\ numerus$
$1.\ x > 0$
$2.\ 3x + 7 > 0$
$3x > -7$
$x > -\dfrac73$
$1\cap2 → x > 0$ . . . . *

$Syarat\ pertidaksamaan$
$2logx \leq log(3x + 7) + 2log\ 2$
$logx^2 \leq log(3x + 7) + log\ 2^2$
$logx^2 \leq log(3x + 7)4$
$x^2 \leq 12x + 28$
$x^2 - 12x - 28 \leq 0$
$(x + 2)(x - 14) \leq 0$
$-2 \leq x \leq 14$ . . . . **


$*\ \cap\ ** → 0 < x \leq 14$
$Jawab:\ C.$

$28.$ Jika diketahui bahwa $^{a^2}log\ b +\ ^{b^2}log\ a = 1$ dimana $a,\ b > 0$ dan $a,\ b \ne 1$, maka nilai $a + b =$ . . . .
$A.\ \dfrac{a^2 + 1}{a}$
$B.\ 2\sqrt{a}$
$C.\ 2a$
$D.\ a^2$
$E.\ a^{1 + \sqrt{2}}$
[Soal SIMAK UI 2011 Matdas]
[Soal dan Pembahasan Logaritma]

$^{a^2}log\ b +\ ^{b^2}log\ a = 1$
$\dfrac12.^alog\ b +\ \dfrac12.^bloga = 1$
$^alog\ b +\ ^blog\ a = 2$
$\dfrac{log\ b}{log\ a} + \dfrac{log\ a}{log\ b} = 2$
$\dfrac{log^2\ a + log^2\ b}{log\ a.log\ b} = 2$
$log^2\ a + log^2\ b = 2log\ a.log\ b$
$log^2\ a -2log\ a.log\ b + log^2\ b = 0$
$(log\ a - log\ b)^2 = 0$
$log\ a - log\ b = 0$
$log\ a = log\ b$
$a = b$

$a + b = a + a = 2a$
$Jawab:\ C.$

$29.$ Jika $x_1\ dan\ x_2$ memenuhi persamaan $(2log\ x - 1)\dfrac{1}{^xlog\ 10} = 10$ maka $x_1x_2 =$ . . . .
$A.\ 5\sqrt{10}$
$B.\ 4\sqrt{10}$
$C.\ 3\sqrt{10}$
$D.\ 2\sqrt{10}$
$E.\ \sqrt{10}$
[UM UGM 2016 Matematika IPA]
[Soal dan Pembahasan Logaritma]

$(2log\ x - 1)\dfrac{1}{^xlog\ 10} = 10$
$(2log\ x - 1)log\ x = 10$
$2log^2\ x - log\ x - 10 = 0$
$(2log\ x - 5)(log\ x + 2) = 0$
$2log\ x - 5 = 0\ atau\ log\ x + 2 = 0$
$log\ x = \dfrac52\ atau\ log\ x = -2$
$x = 10^{\frac52}\ atau\ x = 10^{-2}$
$x_1.x_2 = 10^{\frac52}.10^{-2}$
$= 10^{\frac12}$
$= \sqrt{10}$
$Jawab:\ E.$

$30.$ Jika $^{\sqrt{5}}log(x - 3y) =\ ^5log\ 2x +\ ^5log\ 2y$ maka $\dfrac{x}{y} =$ . . . .
$A.\ \dfrac19$
$B.\ 1$
$C.\ 5$
$D.\ 9$
$E.\ 18$
[Soal UM UGM 2016 Matdas]
[Soal dan Pembahasan Logaritma]

$^{\sqrt{5}}log(x - 3y) =\ ^5log\ 2x +\ ^5log\ 2y$
$\dfrac{1}{\frac12}.^5log(x - 3y) =\ ^5log\ 2x.2y$
$2^5log(x - 3y) =\ ^5log4xy$
$^5log(x - 3y)^2 =\ ^5log4xy$
$(x - 3y)^2 = 4xy$
$x^2 - 6xy + 9y^2 = 4xy$
$x^2 - 10xy + 9y^2 = 0$
$(x - 9y)(x - y) = 0$
$x = 9y\ atau\ x = y$
$x = y$ tidak memenuhi syarat numerus !
$x = 9y → \dfrac{x}{y} = 9$
$Jawab:\ D.$

$31.$ Jika $^4log\ 6 = m + 1,\ ^9log\ 8 =$ . . . .
$A.\ \dfrac{3}{2m + 4}$
$B.\ \dfrac{3}{4m + 2}$
$C.\ \dfrac{3}{4m - 2}$
$D.\ \dfrac{3}{2m - 4}$
$A.\ \dfrac{3}{2m + 2}$
[Soal SPMB 2006 Matdas]
[Soal dan Pembahasan Logaritma]

$^4log\ 6 = m + 1$
$^{2^2}log\ 2.3 = m + 1$
$\dfrac12.^2log\ 2.3 = m + 1$
Semua dikali $2\ !$
$^2log\ 2 +\ ^2log\ 3 = 2m + 2$
$1 +\ ^2log\ 3 = 2m + 2$
$^2log\ 3 = 2m + 1$
$^3log\ 2 = \dfrac{1}{2m + 1}$ . . . . *

$^9log\ 8 =\ ^{3^2}log\ 2^3$
$= \dfrac32.\ ^3log\ 2$ . . . . **
$Susi\ pers\ *\ ke\ pers\ **$
$= \dfrac32.\ \dfrac{1}{2m + 1}$
$= \dfrac{3}{4m + 2}$
$Jawab:\ B.$

$32.$ Jika $x_1\ dan\ x_2$ adalah akar-akar persamaan $(5 - 2log\ x)log\ x = log\ 1.000$, maka $x_1^2 + x_2^2 =$ . . . .
$A.\ 0$
$B.\ 10$
$C.\ 100$
$D.\ 1.000$
$E.\ 1.100$
[Soal SPMB 2007 Matdas]
[Soal dan Pembahasan Logaritma]

$(5 - 2log\ x)log\ x = log\ 1.000$
$5log\ x - 2log^2\ x = 3$
$-2log^2\ x + 5log\ x - 3 = 0$
$2log^2\ x - 5log\ x + 3 = 0$
$(2log\ x - 3)(log\ x - 1) = 0$
$log\ x = \dfrac32\ atau\ log\ x = 1$
$x = 10^{\frac32}\ atau\ x = 10^1$
$x_1 = \sqrt{1.000}\ atau\ x_2 = 10$
$x_1^2 + x_2^2 = (\sqrt{1.000})^2 + 10^2$
$= 1.000 + 100$
$= 1.100$
$Jawab:\ E.$

$33.$ Jika $^7log\ 2 = a$ dan $^2log\ 3 = b,$ maka $^6log\ 98 =$ . . . .
$A.\ \dfrac{a}{a + b}$
$B.\ \dfrac{a + 2}{b + 1}$
$C.\ \dfrac{a + 2}{a(b + 1)}$
$D.\ \dfrac{a + 1}{b + 2}$
$E.\ \dfrac{a + 2}{b(a + 1)}$
[Soal SNMPTN 2008 Matdas]
[Soal dan Pembahasan Logaritma]

$^7log\ 2 = a →\ ^2log\ 7 = \dfrac{1}{a}$
$^2log\ 3 = b →\ ^3log\ 2 = \dfrac{1}{b}$

$^6log\ 98 = \dfrac{log\ 98}{log\ 6}$
$= \dfrac{log\ 7^2.2}{log\ 3.2}$
$= \dfrac{log\ 7^2 + log\ 2}{log\ 3 + log\ 2}$
$Tambahkan\ bilangan\ pokok\ 2\ !$
$= \dfrac{2.\ ^2log\ 7 +\ ^2log\ 2}{^2log\ 3 +\ ^2log\ 2}$
$= \dfrac{2.\dfrac{1}{a} + 1}{b + 1}$

$= \dfrac{\dfrac{2 + a}{a}}{b + 1}$

$= \dfrac{a + 2}{a(b + 1)}$
$Jawab:\ C.$

$34.$ Jika $2^x = a$ dan $2^y = b$ dengan $x,\ y > 0$ maka $\dfrac{2x + 3y}{x + 2y} =$ . . . .
$A.\ \dfrac35$
$B.\ \dfrac53$
$C.\ 1 +\ ^{ab}log\ ab^2$
$D.\ 1 +\ ^{ab}log\ a^2b$
$E.\ 1 +\ ^{ab^2}log\ ab$
[Soal UM UGM 2009 Matdas]
[Soal dan Pembahasan Logaritma]

$2^x = a →\ ^2log\ a = x$
$2^y = b →\ ^2log\ b = y$
$\dfrac{2x + 3y}{x + 2y} = \dfrac{2.\ ^2log\ a + 3.\ ^2log\ b}{^2log\ a + 2.\ ^2log\ b}$
$= \dfrac{^2log\ a^2 +\ ^2log\ b^3}{^2log\ a +\ ^2log\ b^2}$
$= \dfrac{^2log\ a^2b^3}{^2log\ ab^2}$
$=\ ^{ab^2}log\ a^2b^3$
$=\ ^{ab^2}log\ ab^2.ab$
$=\ ^{ab^2}log\ ab^2 +\ ^{ab^2}log\ ab$
$=\ 1 +\ ^{ab^2}log\ ab$
$Jawab:\ E.$

$35.$ Jika $^{81}log\ \dfrac{1}{x} =\ ^xlog\ \dfrac{1}{y} =\ ^ylog\ \dfrac{1}{81}$, maka $2x - 3y =$ . . . .
$A.\ -162$
$B.\ -81$
$C.\ 0$
$D.\ 81$
$E.\ 162$
[Soal SPMB 2006 Matematika IPA]
[Soal dan Pembahasan Logaritma]

$\Leftrightarrow$ $\ ^{81}log\ \dfrac{1}{x} =\ ^xlog\ \dfrac{1}{y} =\ ^ylog\ \dfrac{1}{81}$
$\Leftrightarrow$ $\ ^{81}log\ x^{-1} =\ ^xlog\ y^{-1} =\ ^ylog\ 81^{-1}$
$\Leftrightarrow$ $\ (-1).^{81}log\ x = (-1).^xlog\ y = (-1).^ylog\ 81$
$\Leftrightarrow$ $\ ^{81}log\ x =\ ^xlog\ y =\ ^ylog\ 81$
$\ Misalkan:$
$\Leftrightarrow$ $\ ^{81}log\ x =\ ^xlog\ y =\ ^ylog\ 81 = p$
$\ Kalikan\ semuanya\ !$
$\Leftrightarrow$ $\ ^{81}log\ x . ^xlog\ y . ^ylog\ 81 = p.p.p$
$\Leftrightarrow$ $\ \dfrac{log\ x}{log\ 81}.\dfrac{log\ y}{log\ x}.\dfrac{log\ 81}{log\ y} = p^3$
$\Leftrightarrow$ $\ 1 = p^3$
$\Leftrightarrow$ $\ p = 1$

$\Leftrightarrow$ $\ ^{81}log\ x = 1 →\ x = 81$
$\Leftrightarrow$ $\ ^ylog\ 81 = 1 → 81 = y^1 → y = 81$
$\Leftrightarrow$ $\ 2x - 3y = 2.81 - 3.81 = (2 - 3).81 = -81$
$\ Jawab:\ B.$

$36.$ Nilai $\left(^alog\ \dfrac{1}{b^2} \right)\left(^blog\ \dfrac{1}{c^2} \right)\left(^clog\ \dfrac{1}{a^3} \right) =$ . . . .
$A.\ -6$
$B.\ -8$
$C.\ -10$
$D.\ -12$
$E.\ -14$
[Soal SNMPTN 2010 Matdas]
[Soal dan Pembahasan Logaritma]

$\left(^alog\ \dfrac{1}{b^2} \right)\left(^blog\ \dfrac{1}{c^2} \right)\left(^clog\ \dfrac{1}{a^3} \right)$ $=\ ^alog\ b^{-2}.^blog\ c^{-2}.^clog\ a^{-3}$
$= (-2).(-2).(-3).^alog\ b.^blog\ c.^clog\ a$
$= -12.\dfrac{log\ b}{log\ a}.\dfrac{log\ c}{log\ b}.\dfrac{log\ a}{log\ c}$
$= -12.1$
$= -12$
$Jawab: D.$

$37.$ Jika $6(3^{40})(^2log\ a) + 3^{41}(^2log\ a) = 3^{43}$ Maka nilai $a$ adalah . . . .
$A.\ \dfrac18$
$B.\ \dfrac14$
$C.\ 4$
$D.\ 8$
$E.\ 16$
[Soal SNMPTN 2011 Matdas]
[Soal dan Pembahasan Logaritma]

$6(3^{40})(^2log\ a) + 3^{41}(^2log\ a) = 3^{43}$
$2.3.3^{40}.(^2log\ a) + 3^{41}.(^2log\ a) = 3^{43}$
$2.3^{41}.(^2log\ a) + 3^{41}.(^2log\ a) = 3^{43}$
$3.3^{41}.(^2log\ a) = 3^{43}$
$3^{42}.(^2log\ a) = 3^{43}$
$^2log\ a = \dfrac{3^{43}}{3^{42}}$
$^2log\ a = 3$
$a = 2^3 = 8$
$Jawab: D.$

$38.$ Jika $x > y > 1$ dan $x^2 + 4y^2 = 12xy$, maka $log\dfrac{(x + 2y)^2}{(x - 2y)^2} =$ . . . .
$A.\ 2$
$B.\ 4$
$C.\ -log\ 2$
$D.\ log\ 2$
$E.\ 2log\ 2$
[Soal UMPTN 2001 Matdas]
[Soal dan Pembahasan Logaritma]

$x^2 + 4y^2 = 12xy$
Kurangkan ruas kiri
dan ruas kanan dengan $4xy$
$x^2 - 4xy + 4y^2 = 12xy - 4xy$
$(x - 2y)^2 = 8xy$ . . . . *

$x^2 + 4y^2 = 12xy$
tambahkan ruas kiri
dan ruas kanan dengan $4xy$
$x^2 + 4x + 4y^2 = 12xy + 4xy$
$(x + 2y)^2 = 16xy$ . . . . **

$log\dfrac{(x + 2y)^2}{(x - 2y)^2} = log\ \dfrac{16xy}{8xy} = log\ 2$
$Jawab:\ D.$

$39.$ Jika $f(x) = \dfrac{^{11}log\ x}{1 - 2^{11}log\ x}$, maka $f(x) + f\left(\dfrac{11}{x}\right),$ adalah . . . .
$A.\ -11$
$B.\ -9$
$C.\ -7$
$D.\ -2$
$E.\ -1$
[Soal UMPTN 1995 Matdas]
[Soal dan Pembahasan Logaritma]

$f(x) + f\left(\dfrac{11}{x}\right) = \dfrac{^{11}log\ x}{1 - 2^{11}log\ x} + \dfrac{^{11}log\ \dfrac{11}{x}}{1 - 2^{11}log\ \dfrac{11}{x}}$
$= \dfrac{^{11}log\ x}{1 - 2^{11}log\ x} + \dfrac{^{11}log\ 11 -\ ^{11}log\ x}{1 - 2(^{11}log\ 11 -\ ^{11}log\ x)}$
$= \dfrac{^{11}log\ x}{1 - 2^{11}log\ x} + \dfrac{1 -\ ^{11}log\ x}{1 - 2 +\ 2^{11}log\ x}$
$= \dfrac{^{11}log\ x}{1 - 2^{11}log\ x} + \dfrac{1 -\ ^{11}log\ x}{-1 +\ 2^{11}log\ x}$
$= \dfrac{^{11}log\ x}{1 - 2^{11}log\ x} + \dfrac{1 -\ ^{11}log\ x}{-(1 -\ 2^{11}log\ x)}$
$= \dfrac{^{11}log\ x}{1 - 2^{11}log\ x} - \dfrac{1 -\ ^{11}log\ x}{(1 -\ 2^{11}log\ x)}$
$= \dfrac{^{11}log\ x - 1 +\ ^{11}log\ x }{1 - 2^{11}log\ x}$
$= \dfrac{-1 + 2^{11}log\ x}{1 - 2^{11}log\ x}$
$= \dfrac{-(1 - 2^{11}log\ x)}{(1 - 2^{11}log\ x)}$
$= -1$
$Jawab:\ E.$

$40.$ Jika $x$ dan $y$ memenuhi persamaan $log(3x^3y^3) = 1 + 2log\ x + log\ y$ maka
$A.\ x + y = \dfrac{10}{3}$
$B.\ \dfrac{x}{y} = \dfrac{10}{3}$
$C.\ xy^2 = \dfrac{10}{3}$
$D.\ x - y = \dfrac{10}{3}$
$E.\ 2x + y = \dfrac{10}{3}$
[Soal SPMB 2005 Matdas]
[Soal dan Pembahasan Logaritma]

$log(3x^3y^3) = 1 + 2log\ x + log\ y$
$log(3x^3y^3) = log\ 10 + log\ x^2 + log\ y$
$log(3x^3y^3) = log\ 10x^2y$
$3x^3y^3 = 10x^2y$
$3xy^2 = 10$
$xy^2 = \dfrac{10}{3}$
$Jawab: C.$

Demikianlah soal dan pembahasan logaritma khusus soal-soal UNBK dan SBMPTN, semoga bermanfaat. Selamat belajar !

Disusun oleh:
Joslin Sibarani
Alumni Teknik Sipil ITB

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