Question

$4^x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2x-1}$ find the value of x

Answer 1

The given equation can be re-written as
$2^{2x}+2^{2x-1}=3^{x-1/2}+3^{x+1/2}$
$2^{2x}(1+\frac{1}{2})=3^x(\frac{1}{\sqrt{\left(3\right)}}+\sqrt{3})$
or ${3.2}^{2x-1}=4.3^{x-1/2}$
or $2^{2x-3}=3^{x-3/2}$
Now by trial, $x=\frac{3}{2}$ is one solution
Now taking logarithm to the base 10, we get
$(2x-3)lo{g}_{10}2=(x-\frac{3}{2})lo{g}_{10}3$
or $(2log2-log3)x=$\dfrac{3}{2}(2log 2 \, - \, log 3)$Thisalsogives$x \, = \, \dfrac{3}{2}$
Hence $x=\frac{3}{2}$is the only solution