Question

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

• A. $0$
• B. $-1$
• C. $3$
• D. None of these

Answer 1

The correct option is A $0$

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

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

So, ${(2^x-3)}^2=(\frac{17}{2}+2^{x-1})2$

$2^{2x}-6\times 2^x+9=17+2^x$

$2^{2x}-7\times 2^x-8=0$

Let $2^x=Z$

So, $Z^2-7x-8=0$

$(Z-8)(Z+1)=0$

$Z=-1$ and $Z=8$

$2^x=-1$ (Not Possible)

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