If log _32,log _3( 2^x - 5 ),log _3( 2^x - (7 / 2) ) are in an arithmetic progression, then the value of x is equal to

Because log _3 2, log _3(2^x-5), log _3(2^x-(7 / 2)) in A.P. \ => 2 log _3(2^x-5) =log _3 2+log _3(2^x-(7 / 2)) \ => (2^x-5)^2 =2(2^x-(7 / 2)) => (2^x-5)^2 =(2^x+1-7) \ => 2^2 x+25-5 * 2^x+1 =2^x+1-7 \ => 2^2 x-12 * 2^x+32=0 Let 2^x=t => t^2-12 t+32=0 => (t-8)(t-4) =0 => 2^x=2^3 or 2^x=2^2 => x=3 or 2 But for $x=2,(2^x-5)

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

$\begin{aligned} & \because \log _3 2, \log _3\left(2^x-5\right), \log _3\left(2^x-\frac{7}{2}\right) \in \text { A.P. } \\ & \Rightarrow 2 \log _3\left(2^x-5\right) =\log _3 2+\log _3\left(2^x-\frac{7}{2}\right) \\ & \Rightarrow\left(2^x-5\right)^2 =2\left(2^x-\frac{7}{2}\right) \Rightarrow\left(2^x-5\right)^2 =\left(2^{x+1}-7\right) \\ & \Rightarrow 2^{2 x}+25-5 \cdot 2^{x+1} =2^{x+1}-7 \\ & \Rightarrow 2^{2 x}-12 \cdot 2^x+32=0 \end{aligned}\\$ Let $2^x=t \Rightarrow t^2-12 t+32=0 \Rightarrow(t-8)(t-4) \\=0 \Rightarrow 2^x=2^3$ or $2^x=2^2 \Rightarrow x=3$ or 2 $\\$ But for $x=2,\left(2^x-5\right)<0$ which is not possible. $\therefore x=3$

Explanation

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