If log _32,log _3( 2^x - 5 ) and log _3( 2^x - (7 / 2) ) are in arithmetic progression, determine the value of x

2 log _3(2^x-5)=log _3 2+log _3(2^x-(7 / 2)) \ Therefore (2^x-5)^2=2 x (2^x-(7 / 2)) Therefore (2^x)^2-10 x 2^x+25=2 x 2^x-7 is a quadratic equation in 2^x . Therefore (2^x)^2-12 x 2^x+32=0 Therefore (2^x-4)(2^x-8)=0 Therefore 2^x=4 or 2^x=8 Therefore x=2 or 3 For x=2, log _3(2^x-5) is not meaningful Therefore x=3 is the real solution.

If \log_{3}2,\log_{3}( 2^{x} - 5 ) and \log_{3}( 2^{... | Sequence & Series (Mathematics)

$2 \log _3\left(2^x-5\right)=\log _3 2+\log _3\left(2^x-\frac{7}{2}\right)$ $\\ \therefore\left(2^x-5\right)^2=2 \times\left(2^x-\frac{7}{2}\right) \\$ $\therefore\left(2^x\right)^2-10 \times 2^x+25=2 \times 2^x-7$ { is a quadratic equation in } $2^x . \\$ $\therefore\left(2^x\right)^2-12 \times 2^x+32=0 \\$ $\therefore\left(2^x-4\right)\left(2^x-8\right)=0 \\$ $\therefore 2^x=4$ { or } $2^x=8 \therefore x=2$ { or } 3 For $x=2, \log _3\left(2^x-5\right)\\$ is not meaningful $\therefore x=3$ is the real solution.

Explanation

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