Solve for x square root of 3x-5 = square root of 2x

$\sqrt{3 x - 5} = \sqrt{2 x}$

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{3 x - 5}}^{2} = {\sqrt{2 x}}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 x - 5}$ as ${(3 x - 5)}^{\frac{1}{2}}$ .

${({(3 x - 5)}^{\frac{1}{2}})}^{2} = {\sqrt{2 x}}^{2}$

Simplify the left side.

Simplify ${({(3 x - 5)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(3 x - 5)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(3 x - 5)}^{\frac{1}{2} \cdot 2} = {\sqrt{2 x}}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(3 x - 5)}^{\frac{1}{2} \cdot 2} = {\sqrt{2 x}}^{2}$

Rewrite the expression.

${(3 x - 5)}^{1} = {\sqrt{2 x}}^{2}$

Simplify.

$3 x - 5 = {\sqrt{2 x}}^{2}$

Simplify the right side.

Rewrite ${\sqrt{2 x}}^{2}$ as $2 x$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x}$ as ${(2 x)}^{\frac{1}{2}}$ .

$3 x - 5 = {({(2 x)}^{\frac{1}{2}})}^{2}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$3 x - 5 = {(2 x)}^{\frac{1}{2} \cdot 2}$

Combine $\frac{1}{2}$ and $2$ .

$3 x - 5 = {(2 x)}^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$3 x - 5 = {(2 x)}^{\frac{2}{2}}$

Rewrite the expression.

$3 x - 5 = {(2 x)}^{1}$

Simplify.

$3 x - 5 = 2 x$

Solve for $x$ .

Move all terms containing $x$ to the left side of the equation.

Subtract $2 x$ from both sides of the equation.

$3 x - 5 - 2 x = 0$

Subtract $2 x$ from $3 x$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$