સમીકરણ $log_7(2^x -1) + log_7(2^x -7) = 1$ ના ઉકેલોની સંખ્યા મેળવો.
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b
\(2^{x}=t\)
\(2^{x}=t\)
\(\log _{7}(t-1)+\log _{7}(x->)=1\)
\(\log [(t-1)(x-7)]=1\)
\((x-1)(t-2)=7\)
\(k^{2}-7 t-x+7=7\)
\(t^{2}-8 t=0\)
\(t_{1}=0\) or \(t=8\)
\(2^{x}=0 \quad\) or \(\quad 2^{x}=8\)
\(2^{x}=2^{3}\)
\(\Rightarrow x=3\)
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