Question

If $(4{)}^{{\log }_{9}3}+(9{)}^{{\log }_{2}4}=(10{)}^{{\log }_{x}83}$, then the value of $x$ is equal to

• A. $2$
• B. $3$
• C. $10$
• D. $30$

Answer 1

Answer: (C)

$10$

$(4{)}^{{\log }_{9}3}+(9{)}^{{\log }_{2}4}=(10{)}^{{\log }_{x}83}$
Taking L.H.S.
$(4{)}^{{\log }_{9}3}+(9{)}^{{\log }_{2}4}$
$=2^{2{\log }_{9}3}+3^{2{\log }_{2}4}$
$=2^{{\log }_{9}9}+3^{2{\log }_2{2}^2}$
$=2^{{\log }_{9}9}+3^{4{\log }_{2}2}$
$=2^1+3^4$ $(\because$ ${\log }_{a}a=1)$
$=2+81=83$
Taking $L.H.S=R.H.S$
$83={10}^{{\log }_{x}83}={83}^{{\log }_{x}10},(\because a^{{\log }_{x}y}=y^{{\log }_{x}a})$
$\Rightarrow {\log }_{x}10=1$
$\therefore x={10}^1=10$