If {log _5}2,,{log _5}({2^x} - 3) | vedclass

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Question

$2 \mathrm{b}=\mathrm{a}+\mathrm{c}$

$\Rightarrow 2 \log _{5}\left(2^{\mathrm{x}}-3\right)=\log _{5} 2+\log _{5}\left(\frac{17}{2}+2^{\mathrm{x}-1}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

$2 \mathrm{b}=\mathrm{a}+\mathrm{c}$

$\Rightarrow 2 \log _{5}\left(2^{\mathrm{x}}-3\right)=\log _{5} 2+\log _{5}\left(\frac{17}{2}+2^{\mathrm{x}-1}\right)$

$\Rightarrow\left(2^{\mathrm{x}}-3\right)^{2}=2\left(\frac{17}{2}+2^{\mathrm{x}-1}\right) \Rightarrow 2^{2 \mathrm{x}}-7.2^{\mathrm{x}}-8=0$

$\Rightarrow \mathrm{x}=3$

If ${\log _5}2,\,{\log _5}({2^x} - 3)$ and ${\log _5}(\frac{{17}}{2} + {2^{x - 1}})$ are in $A.P.$ then the value of $x$ is :-

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