SBMPTN Zone : Persamaan Logaritma

Berikut ini adalah kumpulan soal mengenai persamaan logaritma Tipe SBMPTN. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

Tipe:

Standar SBMPTN HOTS

No. 1

Jika x_1 dan x_2 memenuhi \left(^{(x-1)}\log4\right)^2=4, maka nilai {x_1+x_2} adalah

1. 3
2. 3\dfrac12
3. 4

4. 4\dfrac12
5. 5

Syarat:
$\begin{array}{rl}x-1& >0\\ x& >1\end{array}$

$\begin{array}{rl}{({}^{(x-1)}\log 4)}^2& =4\\ {}^{(x-1)}\log 4& =\pm 2\end{array}$

$\begin{array}{rl}{}^{(x-1)}\log 4& =2\\ (x-1{)}^2& =4\\ x^2-2x+1& =4\\ x^2-2x-3& =0\\ (x+1)(x-3)& =0\end{array}$ x=-1 (PM) atau \boxed{x=3}

$\begin{array}{rl}{}^{(x-1)}\log 4& =-2\\ (x-1{)}^{-2}& =4\\ {\displaystyle \frac{1}{(x-1{)}^2}}& =4\\ (x-1{)}^2& ={\displaystyle \frac{1}{4}}\\ x^2-2x+1& ={\displaystyle \frac{1}{4}}\\ 4x^2-8x+4& =1\\ 4x^2-8x+3& =0\\ (2x-1)(2x-3)& =0\end{array}$ x=\dfrac12 (PM) atau \boxed{x=\dfrac32=1\dfrac12}

$\begin{array}{rl}{x}_1+x_2& =3+1{\displaystyle \frac{1}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 4{\displaystyle \frac{1}{2}}}}}}\end{array}$

Jadi, {x_1+x_2=4\dfrac12}.
JAWAB: D

No. 2

Jika x_1 dan x_2 memenuhi \left({^{x-2}\log}9\right)^2=4, maka nilai x_1+x_2 adalah ....

Ganesha Operation

$\begin{array}{rl}{\left({}^{x-2}\log 9\right)}^2& =4\\ {}^{x-2}\log 9& =\pm 2\\ x-2& =9^{\pm \frac{1}{2}}\end{array}$

• x_1-2=9^{\frac12}
  $\begin{array}{rl}{x}_1& =2+3\\ & =5\end{array}$
• x_2-2=9^{-\frac12}
  $\begin{array}{rl}{x}_2& =2+{\displaystyle \frac{1}{3}}\\ & =2{\displaystyle \frac{1}{3}}\end{array}$
  

x_1+x_2=5+2\dfrac13=7\dfrac13

Jadi, nilai x_1+x_2 adalah 7\dfrac13.

No. 3

Jika x_1 dan x_2 memenuhi {\left({^{27}\negthinspace\log}\dfrac1{x+1}\right)^2=\dfrac19}, maka nilai x_1x_2 adalah ....

1. \dfrac53
2. \dfrac43
3. \dfrac13

4. -\dfrac23
5. -\dfrac43

SBMPTN 2018 Kode 517

$\begin{array}{rl}{\left({}^{27}\log {\displaystyle \frac{1}{x+1}}\right)}^2& ={\displaystyle \frac{1}{9}}\\ {}^{27}\log {\displaystyle \frac{1}{x+1}}& =\pm {\displaystyle \frac{1}{3}}\\ {\displaystyle \frac{1}{x+1}}& ={27}^{\pm \frac{1}{3}}\\ & ={\left(3^3\right)}^{\pm \frac{1}{3}}\\ & =3^{\pm 1}\\ x+1& ={\displaystyle \frac{1}{3^{\pm 1}}}\\ x& =-1+{\displaystyle \frac{1}{3^{\pm 1}}}\end{array}$

$\begin{array}{rl}{x}_1& =-1+{\displaystyle \frac{1}{3}}\\ & =-{\displaystyle \frac{2}{3}}\end{array}$

$\begin{array}{rl}{x}_2& =-1+{\displaystyle \frac{1}{3^{-1}}}\\ & =-1+3\\ & =2\end{array}$

$\begin{array}{rl}{x}_1{x}_2& =(-{\displaystyle \frac{2}{3}})(2)\\ & =-{\displaystyle \frac{4}{3}}\end{array}$

Jadi, nilai x_1x_2 adalah \dfrac43.
JAWAB: E

No. 4

Jika ^3\negthinspace\log p+{^9\negthinspace\log q} = 5 dan ^9\negthinspace\log q^8 +{^3\negthinspace\log p^5} = 11, maka nilai dari ^q\negthinspace\log p^2 adalah ....

1. 6\ ^3\negthinspace\log p
2. 6\ ^3\negthinspace\log q
3. 3\ ^3\negthinspace\log p

4. -3\ ^3\negthinspace\log q
5. -3\ ^3\negthinspace\log q

http://www.learncy.net/problem/166/

$\begin{array}{rl}{}^9\log q^8+{}^3\log p^5& =11\\ 8{}^9\log q+5{}^3\log p& =11\\ 5{}^9\log q+5{}^3\log p& =25-\\ 3{}^9\log q& =-14\\ {}^9\log q& =-{\displaystyle \frac{14}{3}}\end{array}$

$\begin{array}{rl}{}^q\log p^2& ={\displaystyle \frac{{}^9\log p^2}{{}^9\log q}}\\ & ={\displaystyle \frac{{}^{3^2}\log p^2}{-{\displaystyle \frac{14}{3}}}}\end{array}$

Jadi, ^q\negthinspace\log p^2=.

No. 5

Jika xy= 90 dan \log x-\log y= 1, maka x-y= ....

1. 27
2. 25
3. -26

4. 19
5. 20

Syarat:

• x\gt0
• y\gt0

$\begin{array}{rl}\log x-\log y& =1\\ \log {\displaystyle \frac{x}{y}}& =\log 10\\ {\displaystyle \frac{x}{y}}& =10\\ x& =10y\end{array}$

$\begin{array}{rl}xy& =90\\ (10y)y& =90\\ 10y^2& =90\\ y^2& =9\\ y& =\boxed{{\displaystyle 3}}\end{array}$

$\begin{array}{rl}x& =10y\\ & =10(3)\\ & =\boxed{{\displaystyle 30}}\end{array}$

$\begin{array}{rl}x-y& =30-3\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 27}}}}\end{array}$

Jadi, x-y=27.
JAWAB: A

No. 6

Jika $\log \left(x^2\right)+\log \left(10x^2\right)+\log \left({10}^2{x}^2\right)+\cdots +\log \left({10}^9{x}^2\right)=55$, maka x= ....

$\begin{array}{rl}\log \left(x^2\right)+\log \left(10x^2\right)+\log \left({10}^2{x}^2\right)+\cdots +\log \left({10}^9{x}^2\right)& =55\\ \log (x^2\cdot 10x^2\cdot {10}^2{x}^2\cdots {10}^9{x}^2)& =55\\ \log \left({10}^{1+2+\cdots +9}{x}^{20}\right)& =55\\ \log \left({10}^{45}{x}^{20}\right)& =55\\ {10}^{45}{x}^{20}& ={10}^{55}\\ x^{20}& ={\displaystyle \frac{{10}^{55}}{{10}^{45}}}\\ & ={10}^{10}\\ x& ={10}^{\frac{10}{20}}\\ & ={10}^{\frac{1}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \sqrt{10}}}}}\end{array}$

Jadi, x=\sqrt{10}.
JAWAB: D

No. 7

Penyelesaian dari (2x)^{1+\log_22x}\geq64x^3 adalah

1. 0\lt x\leq\dfrac14
2. \dfrac14\leq x\leq4
3. x\leq\dfrac14 atau x\geq4

4. 0\lt x\leq\dfrac14 atau x\geq4
5. \dfrac14\leq x\leq2 atau x\gt4

$\begin{array}{rl}(2x{)}^{1+{\log }_{2}2x}& \ge 64x^3\\ {\log }_2((2x{)}^{1+{\log }_{2}2x})& \ge {\log }_{2}64x^3\\ (1+{\log }_{2}2x){\log }_{2}2x& \ge {\log }_2(8\cdot 8x^3)\\ {\log }_{2}2x+{{\log }_2}^{2}2x& \ge {\log }_{2}8+{\log }_{2}8x^3\\ {{\log }_2}^{2}2x+{\log }_{2}2x& \ge 3+{\log }_2(2x{)}^3\\ {{\log }_2}^{2}2x+{\log }_{2}2x& \ge 3+3{\log }_{2}2x\end{array}$
Misal \log_22x=p
$\begin{array}{rl}{p}^2+p& \ge 3+3p\\ p^2-2p-3& \ge 0\\ (p+1)(p-3)& \ge 0\end{array}$

p\leq-1	atau	p\geq3
\log_22x\leq-1	atau	\log_22x\geq3
2x\leq2^{-1}	atau	2x\geq2^3
2x\leq\dfrac12	atau	2x\geq8
x\leq\dfrac14	atau	x\geq4

Syarat:

• 2x\gt0
  x\gt0
• 2x\neq1
  x\neq\dfrac12

0\lt x\leq\dfrac14 atau x\geq4

Jadi, 0\lt x\leq\dfrac14 atau x\geq4.
JAWAB: D

No. 8

Jika x memenuhi persamaan {{^5\negmedspace\log 5x} + {^4\negmedspace\log 4x} = {^{25}\negmedspace\log 25x^2}} maka nilai {^x\negmedspace\log 4} =

1. \dfrac12
2. -2
3. 2

4. 1
5. -1

SKMP September 2020

$\begin{array}{rl}{}^5\log 5x+{}^4\log 4x& ={}^{25}\log 25x^2\\ {}^5\log 5x+{}^4\log 4+{}^4\log x& ={}^{5^2}\log (5x{)}^2\\ {}^5\log 5x+1+{}^4\log x& ={}^5\log 5x\\ 1+{}^4\log x& =0\\ {}^4\log x& =\boxed{{\displaystyle \boxed{{\displaystyle -1}}}}\end{array}$

Jadi, {^x\negmedspace\log 4} =-1.
JAWAB: E

No. 9

Jika {{^4\negmedspace\log \sqrt{x}}+ {^2\negmedspace\log y}={^4\negmedspace\log z^2}}, maka z^2 =

1. x\sqrt{y}
2. \sqrt{x}y
3. \sqrt{x}y^2

4. x^2\sqrt{y}
5. xy

SKMP Juli 2020

$\begin{array}{rl}{}^4\log \sqrt{x}+{}^2\log y& ={}^4\log z^2\\ {}^4\log \sqrt{x}+{}^{2^2}\log y^2& ={}^4\log z^2\\ {}^4\log \sqrt{x}+{}^4\log y^2& ={}^4\log z^2\\ {}^4\log \sqrt{x}{y}^2& ={}^4\log z^2\\ \sqrt{x}{y}^2& =z^2\\ z^2& =\boxed{{\displaystyle \boxed{{\displaystyle \sqrt{x}{y}^2}}}}\end{array}$

Jadi, z^2 =.\sqrt{x}y^2
JAWAB: C