Question

$a=\frac{2^{x-1}}{2^{-x-2}}$ and $b=\frac{2^{-x}}{2^{x+1}}$, find the value of $x$.

Answer 1

We have,

$a=\frac{2^{x-1}}{2^{-x-2}}$

$a=2^{x-1}\times 2^{-(-x-2)}$

$a=2^{x-1+x+2}$

$a=2^{2x+1}$

$2x+1={\log }_{2}a$

$2x={\log }_{2}a-1$

$x=\frac{1}{2}({\log }_{2}a-1)$

Since,

$b=\frac{2^{-x}}{2^{x+1}}$

$b=2^{-x}\times 2^{-(x+1)}$

$b=2^{-x-x-1}$

$b=2^{-2x-1}$

$-2x-1={\log }_{2}b$

$-2x=1+{\log }_{2}b$

$x=-\frac{1}{2}(1+{\log }_{2}b)$

Hence, the values of x are $\frac{1}{2}({\log }_{2}a-1)$ and $-\frac{1}{2}(1+{\log }_{2}b)$.