Question

if logx^2(9/16)=-1/2 the value of the base is??

Answer 1

Given $\log_{x^2}(\frac{9}{16} )=-\frac{1}{2}$.

Rearranging the above equation,

$(x^2)^{-\frac{1}{2}}=\frac{9}{16} \\ x^{-1}=\frac{9}{16} \\ x=\frac{16}{9}$

The base of the logarithm is

$x^2=(\frac{16}{9})^2\\ x^2=\frac{256}{81}$

Answer 2

Given:

$\bold{\log x^2 (\frac{9}{16})=-\frac{1}{2}}$

To find:

base value is=?

Solution:

The given equation is solve by two methods which can be described as follows:

$\bold{\to \log x^2 (\frac{9}{16})=-\frac{1}{2}}$

first arrange the value and remove the log we get:

$\to (x^2)^(\frac{-1}{2})=\frac{9}{16}\\\\\to x^{-1}=\frac{9}{16}\\\\\to \frac{1}{x}= \frac{9}{16}\\\\\to x=\frac{16}{9}$

square the above value we get:

$\to x^2= (\frac{16}{9})^2\\\\\to x^2= (\frac{265}{81})\\\\\to x^2= (3.16)$

Or

To rearrange the give value we get the which is in attachment so, please find it: