Question

If $p^{1-2x}\times \sqrt{p}=q^{1-2x}\times \sqrt{q};p\ne q$., then the value of $x$ is $\frac{3}{m}$. The value of $m$ is:

Answer 1

As given $p^{1-2x}\times \sqrt{p}=q^{1-2x}\times \sqrt{q};p\ne q$.
$\Rightarrow p^{1-2x}\times p^{\frac{1}{2}}=q^{1-2x}\times q^{\frac{1}{2}}$
$\Rightarrow p^{1-2x+\frac{1}{2}}=q^{1-2x+\frac{1}{2}}$
Now solving x for p
$\Rightarrow =1-2x+\frac{1}{2}=0$
$\Rightarrow 2-4x+1=0$
$\Rightarrow 3-4x=0$
$\Rightarrow 4x=3$
$\Rightarrow x=\frac{3}{4}$
Now solving x for q
$\Rightarrow =1-2x+\frac{1}{2}=0$
$\Rightarrow 2-4x+1=0$
$\Rightarrow 3-4x=0$
$\Rightarrow 4x=3$
$\Rightarrow x=\frac{3}{4}$