Question

For a real number $x$ , if $\frac{1}{2},\frac{log_3(2^x-9)}{log_34}$, and $\frac{log_5\bigg(2^x+\frac{17}{2}\bigg)}{log_54}$ are in an arithmetic progression, then the common difference is

• A. $log_4\bigg(\frac{23}{2}\bigg)$
• B. $log_4\bigg(\frac{3}{2}\bigg)$
• C. $log_47$
• D. $log_4\bigg(\frac{7}{2}\bigg)$

Answer 1

Answer: (D)

$log_4\bigg(\frac{7}{2}\bigg)$

Answer 2

$\frac{log_3(2^x-9)}{log_34}$ can be written as $log_4(2^x-9)$, and $\frac{log_5\bigg(2^x+\frac{17}{2}\bigg)}{log_54}$ can be written as $log_4\bigg(2^x+\frac{17}{2}\bigg)$
Hence, $2log_4(2^x-9)=\frac12+log_4\bigg(2^x+\frac{17}{2}\bigg)$
$\frac12$ can be written as $log_42$
Therefore,
$2log_4(2^x-9)=log_4\bigg(2^x+\frac{17}{2}\bigg)$

$log_4(2^x-9)^2=log_4\bigg(2^x+\frac{17}{2}\bigg)$

$(2^x-9)^2=2(2^x+\frac{17}{2})$
$2^{2x}-18.2^x+81=2.2^x+17$
$2.2^{2x}-20.2^x+64=0$
$2.2^{2x}-16.2^x-4.2^x+64=0$
$2^x(2^x-16)-4(2^x-16)=0$
$(2^x-4)(2^x-16)=0$
The values of $2^x$ cant be 4 (log will be undefined), which implies The value of $2^x$ is 16.
Therefore, the common difference,
$=log_4 (2^x — 9) — log_42$
$=log_4 7 — log_4 2 = log_4 (\frac72)$

So, the correct option is (D): $log_4\bigg(\frac{7}{2}\bigg)$.