Question

solve 12th one fast....​

Answer 1

$\displaystyle\longrightarrow\sf {(2\log 5+\log 4^2+3\log 2)-2=m\log 2}$

$\quad$

$\displaystyle\longrightarrow\sf {(\log 5^2+\log 4^2+\log 2^3)-\log 2^m=2}$

$\quad$

$\displaystyle\longrightarrow\sf {\log\left (\dfrac {5^2\times4^2\times 2^3}{2^m}\right)=2}$

$\quad$

$\displaystyle\longrightarrow\sf {\log\left (\dfrac {25\times16\times 8}{2^m}\right)=2}$

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {25\times16\times 8}{2^m}=10^2}$

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {25\times16\times 8}{100}=2^m}$

$\quad$

$\displaystyle\longrightarrow\sf {2^m=32}$

$\quad$

$\displaystyle\Longrightarrow\sf {\underline {\underline {m=5}}}$