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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. $0$
in Quantitative Aptitude 9.1k points 21 371 778
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1 Answer

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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, $log_{10}10=1$}$

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

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

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

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

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

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

$\implies$ $x=41$

$\textrm{Option b is correct.}$
3.2k points 4 10 59

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