Lakshman Patel RJIT
asked
in Quantitative Aptitude
Apr 3, 2020
recategorized
Nov 8, 2020
by Krithiga2101

250 views
2 votes

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

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

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

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

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

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

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

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

$\implies$ $x=41$

$\textrm{Option b is correct.}$

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

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

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

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

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

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

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

$\implies$ $x=41$

$\textrm{Option b is correct.}$