Search any question & find its solution
Question:
Answered & Verified by Expert
If $\log _3 2, \log _3\left(2^x-5\right)$ and $\log _3\left(2^x-\frac{7}{2}\right)$ are in A.P., then $x$ is equal to
Options:
Solution:
2836 Upvotes
Verified Answer
The correct answer is:
None of these
$\log _3 2, \log _3\left(2^x-5\right)$ and $\log _3\left(2^x-\frac{7}{2}\right)$ are in A.P.
$\Rightarrow \quad 2 \log _3\left(2^x-5\right)=\log _3$ $\left[(2)\left(2^x-\frac{7}{2}\right)\right]$
$\Rightarrow\left(2^x-5\right)^2=2^{x+1}-7 \Rightarrow 2^{2 x}-12 \cdot 2^x-32=0$
$\Rightarrow \quad x=2,3$
But $x=2$ does not hold, hence $x=3$.
$\Rightarrow \quad 2 \log _3\left(2^x-5\right)=\log _3$ $\left[(2)\left(2^x-\frac{7}{2}\right)\right]$
$\Rightarrow\left(2^x-5\right)^2=2^{x+1}-7 \Rightarrow 2^{2 x}-12 \cdot 2^x-32=0$
$\Rightarrow \quad x=2,3$
But $x=2$ does not hold, hence $x=3$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.