Question

Solve the following equation for $x$:

$\frac{6}{5}{a}^{({\log }_{a}x\cdot {\log }_{10}a\cdot {\log }_{a}5)}-3^{{\log }_{10}\left(\frac{x}{10}\right)}=9^{{\log }_{100}x+{\log }_{4}2}$

• A. $x=10$
• B. $x=100$
• C. $x=2$
• D. $x=4$

Answer 1

The correct option is B $x=100$

Let $A={\log }_{a}x\cdot {\log }_{10}a\cdot {\log }_{a}5$

$={\log }_{10}x\cdot {\log }_{a}5$

$={\log }_a(5{)}^{{\log }_{10}x}$

$\therefore a^A=5^{{\log }_{10}x}=5^{\lambda }$ (say ${\log }_{10}x=\lambda$)

&

Let $B={\log }_{10}\left(\frac{x}{10}\right)={\log }_{10}x-1=\lambda -1$

& $C={\log }_{100}x+{\log }_{4}2$

$={\log }_{{10}^2}{x}^1+{\log }_{2^2}{2}^1$

$=\frac{1}{2}{\log }_{10}x+\frac{1}{2}=\frac{\lambda +1}{2}$

$\therefore 9^c=9^{\frac{\lambda +1}{2}}=3^{\lambda +1}={3.3}^{\lambda }$

According to question

$\frac{6}{5}\cdot 5^{\lambda }-\frac{3^{\lambda }}{3}={3.3}^{\lambda }$

$⟹{6.5}^{\lambda -1}=3^{\lambda }(\frac{1}{3}+3)$

${6.5}^{\lambda -1}=3^{\lambda -1}\left(10\right)$

$⟹5^{\lambda -2}=3^{\lambda -2}$

Which is possible only when

$\lambda =2⟹{\log }_{10}x=2⟹x={10}^2=100$