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Given that: $log_{10}(2^x+x-41)=x(1-log_{10}5)$

 $\because log_aa=1$ ;We can use this property here, $log_{10}10=1$

$\implies log_{10}(2^x+x-41)=x(log_{10}10-log_{10}5)$

$\implies log_{10}(2^x+x-41)=x(log_{10}(\frac{10}{5})) (\because log_ax-log_ay=log_a(\frac{x}{y}))$

$\implies log_{10}(2^x+x-41)=x(log_{10}2)$

$\implies log_{10}(2^x+x-41)=log_{10}2^x (\because log_ax^n=n*log_ax)$

compare both sides we get:

$\implies 2^x+x-41=2^x$

$\implies x=41$

Option (B) is correct.
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