Question

If ${log}_{5}2,{log}_5(2^x-3)and{log}_5(\frac{17}{2}+2^{x-1})$ are in AP. Then the value of x is

• A. $0$
• B. $-1$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

Given,

${\log }_{5}2,{\log }_5(2^x-3),{\log }_5(\frac{17}{2}+2^{x-1})$ are in AP.

$\Rightarrow {\log }_5(2^x-3)-{\log }_{5}2={\log }_5(\frac{17}{2}+2^{x-1})-{\log }_5(2^x-3)$

${\log }_5\left(\frac{2^x-3}{2}\right)={\log }_5\left(\frac{\frac{17}{2}+2^{x-1}}{2^x-3}\right)$

$\frac{2^x-3}{2}=\frac{\frac{17}{2}+2^{x-1}}{2^x-3}$

$(2^x-3{)}^2=17+2^x$

Let $2^x=u$

${(u-3)}^2=17+u$

$u^2-6u+9=17+u$

$u^2-7u-8=0$

$u=8,u=-1$

$2^x=8\Rightarrow x=3$