Solve for b (6b)^(1/2)=(8-2b)^(1/2)

${(6 b)}^{\frac{1}{2}} = {(8 - 2 b)}^{\frac{1}{2}}$

Since the exponents are equal, the bases of the exponents on both sides of the equation must be equal.

$6 b = 8 - 2 b$

Solve for $b$ .

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

Add $2 b$ to both sides of the equation.

$6 b + 2 b = 8$

Add $6 b$ and $2 b$ .

$8 b = 8$

Divide each term in $8 b = 8$ by $8$ and simplify.

Divide each term in $8 b = 8$ by $8$ .

$\frac{8 b}{8} = \frac{8}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 b}{8} = \frac{8}{8}$

Divide $b$ by $1$ .

$b = \frac{8}{8}$

Simplify the right side.

Divide $8$ by $8$ .

$b = 1$