Question

Find the value of $x$ if ${25}^x\sqrt[x]{x{5}^{x-1}}=1$

Answer 1

${25}^x\sqrt[x]{x{5}^{x-1}}=1$

$\Rightarrow 5^{2x}{\left(5^{x-1}\right)}^{\frac{1}{x}}=1$

$\Rightarrow 5^{2x}\left(5^{1-\frac{1}{x}}\right)=1$

$\Rightarrow 5^{2x+1-\frac{1}{x}}=5^0$

Since bases are the same we can equate the powers.

$2x+1-\frac{1}{x}=0$

$\Rightarrow 2x^2+x-1=0$

$\Rightarrow 2x^2+2x-x-1=0$

$\Rightarrow 2x(x+1)-1(x+1)=0$

$\Rightarrow (x+1)(2x-1)=0$

$\therefore x=-1,\frac{1}{2}$