Question

If $\log {}_5^2,\log {}_5^{(2^x-3)}\&\log {}_5^{(\frac{17}{2}+2^{x-1})}$ are in $A.P$. Then the value of $x$ is

• A. $0$
• B. $-1$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

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

Therefore,

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

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

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

$\Rightarrow {\left(2^x\right)}^2+9-6.2^x=17+2^x$

$\Rightarrow {\left(2^x\right)}^2-7\left(2^x\right)-8=0$

$\Rightarrow (2^x-8)(2^x+1)=0$

$\because 2^x\nless 0$

$\therefore 2^x=8$

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