If $\log_x(\frac{9}{16})=-\frac{1}{2}$,then x is equal to :
$-\dfrac{3}{4}$
$\dfrac{3}{4}$
$\dfrac{81}{256}$
$\dfrac{256}{81}$
Explanation:
$\log_(\dfrac{9}{16})$=$-\dfrac{1}{2}$
=>$x^{\dfrac{-1}{2}}$=$\dfrac{9}{16}$
=>$\dfrac{1}{\sqrt{x}}$=$\dfrac{9}{16}$
=>$\dfrac{1}{\sqrt{x}}$=$\dfrac{16}{9}$
=>$\sqrt{x}$=$\dfrac{16}{9}$
=>x=$(\dfrac{16}{9})^{2}$
=>x=$\dfrac{256}{81}$
$\log_(\dfrac{9}{16})$=$-\dfrac{1}{2}$
=>$x^{\dfrac{-1}{2}}$=$\dfrac{9}{16}$
=>$\dfrac{1}{\sqrt{x}}$=$\dfrac{9}{16}$
=>$\dfrac{1}{\sqrt{x}}$=$\dfrac{16}{9}$
=>$\sqrt{x}$=$\dfrac{16}{9}$
=>x=$(\dfrac{16}{9})^{2}$
=>x=$\dfrac{256}{81}$