# NIELIT 2019 Feb Scientist D - Section D: 30

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Find the value of $x$ satisfying : $\log_{10} \left (2^{x}+x-41 \right)=x \left (1-\log_{10}5 \right)$

1. $40$
2. $41$
3. $-41$
4. $0$

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

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

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

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

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

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

$\textrm{compair both side we get:}$

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

$\implies$ $x=41$

$\textrm{Option b is correct.}$
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