Question

9*9^(1/(3x-1))=9^x, Solve for x

Answer 1

$\underline{\underline{\mathbb{FORMULAS :}}}$

$1.\:x^{a}*x^{b}=x^{a+b}$

$2.\:x^{p}=x^{q} \implies p= q$

$\underline{\underline{\mathbb{SOLUTION :}}}$

$\textsf{Given that,}\: 9*9^{\frac{1}{3x-1}}=9^{x}$

$\to 9^{1}*9^{\frac{1}{3x-1}}=9^{x}$

$\to 9^{1+\frac{1}{3x-1}}=9^{x}$

$\boxed{\textsf{since}\:x^{a}*x^{b}=x^{a+b}}$

$\to 1+\frac{1}{3x-1}=x$

$\boxed{\textsf{where}\:x^{p}=x^{q} \implies p= q}$

$\to \frac{3x-1+1}{3x-1}=x$

$\to \frac{3x}{3x-1}=x$

$\to 3x = x(3x-1)$

$\to 3x = 3x^{2}-x$

$\to 3x^{2}-x-3x=0$

$\to 3x^{2}-4x=0$

$\to x(3x-4)=0$

$\textsf{Either x = 0 or, 3x - 4 = 0}$

$\implies \boxed{\bold{x = 0, \frac{4}{3}}}$

$\textsf{which is the required}\:{\textsf{sol}}^{\textsf{n}}.$