Exercise Zone : Persamaan Logaritma

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

Tipe :

Standar SBMPTN HOTS

No. 1

\log_4\left(\log_2x\right)+\log_2\left(\log_4x\right)=2

SUMBER

$\begin{array}{rl}{\log }_4({\log }_{2}x)+{\log }_2({\log }_{4}x)& =2\\ {\log }_4({\log }_{2}x)+{\log }_{2^2}{({\log }_{2^2}x)}^2& ={\log }_4{4}^2\\ {\log }_4({\log }_{2}x)+{\log }_4{({\displaystyle \frac{1}{2}}{\log }_{2}x)}^2& ={\log }_{4}16\\ {\log }_4({\log }_{2}x)+{\log }_4\left({\displaystyle \frac{1}{4}}{({\log }_{2}x)}^2\right)& ={\log }_{4}16\\ {\log }_4({\log }_{2}x\cdot {\displaystyle \frac{1}{4}}{({\log }_{2}x)}^2)& ={\log }_{4}16\\ {\displaystyle \frac{1}{4}}{({\log }_{2}x)}^3& =16\\ {({\log }_{2}x)}^3& =64\\ {\log }_{2}x& =4\\ x& =2^4\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 16}}}}\end{array}$

Jadi, x=16.

No. 2

Tentukan nilai x jika {{^2\negmedspace\log(3x-2)}+{^2\negmedspace\log9}-{^2\negmedspace\log x}=3}

Grup Telegram

$\begin{array}{rl}{}^2\log (3x-2)+{}^2\log 9-{}^2\log x& =3\\ {}^2\log {\displaystyle \frac{(3x-2)9}{x}}& ={}^2\log 2^3\\ {}^2\log {\displaystyle \frac{27x-18}{x}}& ={}^2\log 8\\ {\displaystyle \frac{27x-18}{x}}& =8\\ 27x-18& =8x\\ 19x& =18\\ x& =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{18}{19}}}}}}\end{array}$

Jadi, x=\dfrac{18}{19}.

No. 3

Nilai x yang memenuhi persamaan {{^{10}\negmedspace\log(2x-5)}={^{10}\negmedspace\log(x+3)}}

Grup Telegram

$\begin{array}{rl}{}^{10}\log (2x-5)& ={}^{10}\log (x+3)\\ 2x-5& =x+3\\ x& =\boxed{{\displaystyle \boxed{{\displaystyle 8}}}}\end{array}$

Jadi, x=8.

No. 4

HP dari persamaan logaritma {{^2\negmedspace\log(2x+3)}-{^2\negmedspace\log(3x-4)}=3} adalah...

Grup Telegram

$\begin{array}{rl}{}^2\log (2x+3)-{}^2\log (3x-4)& =3\\ {}^2\log {\displaystyle \frac{2x+3}{3x-4}}& =3\\ {\displaystyle \frac{2x+3}{3x-4}}& =2^3\\ {\displaystyle \frac{2x+3}{3x-4}}& =8\\ {\displaystyle \frac{2x+3}{3x-4}}-8& =0\\ {\displaystyle \frac{2x+3-8(3x-4)}{3x-4}}& =0\\ {\displaystyle \frac{2x+3-24x+32}{3x-4}}& =0\\ {\displaystyle \frac{-22x+35}{3x-4}}& =0\\ -22x+35& =0\\ -22x& =-35\\ x& ={\displaystyle \frac{-35}{-22}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{35}{22}}}}}}\end{array}$

Jadi, Hp = \left\{\dfrac{35}{22}\right\}.

No. 4

Tentukan nilai x jika {{^3\negmedspace\log x}