Solve for b 16ab^2=25

$16 a b^{2} = 25$

Step 1

Divide each term in $16 a b^{2} = 25$ by $16 a$ and simplify.

Divide each term in $16 a b^{2} = 25$ by $16 a$ .

$\frac{16 a b^{2}}{16 a} = \frac{25}{16 a}$

Simplify the left side.

Cancel the common factor of $16$ .

Cancel the common factor.

$\frac{16 a b^{2}}{16 a} = \frac{25}{16 a}$

Rewrite the expression.

$\frac{a b^{2}}{a} = \frac{25}{16 a}$

Cancel the common factor of $a$ .

Cancel the common factor.

$\frac{a b^{2}}{a} = \frac{25}{16 a}$

Divide $b^{2}$ by $1$ .

$b^{2} = \frac{25}{16 a}$

Step 2

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$b = \pm \sqrt{\frac{25}{16 a}}$

Step 3

Simplify $\pm \sqrt{\frac{25}{16 a}}$ .

Rewrite $\frac{25}{16 a}$ as ${(\frac{5}{2^{2}})}^{2} \frac{1}{a}$ .

Factor the perfect power $5^{2}$ out of $25$ .

$b = \pm \sqrt{\frac{5^{2} \cdot 1}{16 a}}$

Factor the perfect power ${(2^{2})}^{2}$ out of $16 a$ .

$b = \pm \sqrt{\frac{5^{2} \cdot 1}{{(2^{2})}^{2} a}}$

Rearrange the fraction $\frac{5^{2} \cdot 1}{{(2^{2})}^{2} a}$ .

$b = \pm \sqrt{{(\frac{5}{2^{2}})}^{2} \frac{1}{a}}$

Pull terms out from under the radical.

$b = \pm \frac{5}{2^{2}} \sqrt{\frac{1}{a}}$

Raise $2$ to the power of $2$ .

$b = \pm \frac{5}{4} \sqrt{\frac{1}{a}}$

Rewrite $\sqrt{\frac{1}{a}}$ as $\frac{\sqrt{1}}{\sqrt{a}}$ .

$b = \pm \frac{5}{4} \cdot \frac{\sqrt{1}}{\sqrt{a}}$

Any root of $1$ is $1$ .

$b = \pm \frac{5}{4} \cdot \frac{1}{\sqrt{a}}$

Multiply $\frac{1}{\sqrt{a}}$ by $\frac{\sqrt{a}}{\sqrt{a}}$ .

$b = \pm \frac{5}{4} (\frac{1}{\sqrt{a}} \cdot \frac{\sqrt{a}}{\sqrt{a}})$

Combine and simplify the denominator.

Multiply $\frac{1}{\sqrt{a}}$ by $\frac{\sqrt{a}}{\sqrt{a}}$ .

$b = \pm \frac{5}{4} \cdot \frac{\sqrt{a}}{\sqrt{a} \sqrt{a}}$

Raise $\sqrt{a}$ to the power of $1$ .

$b = \pm \frac{5}{4} \cdot \frac{\sqrt{a}}{{\sqrt{a}}^{1} \sqrt{a}}$

Raise $\sqrt{a}$ to the power of $1$ .

$b = \pm \frac{5}{4} \cdot \frac{\sqrt{a}}{{\sqrt{a}}^{1} {\sqrt{a}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$b = \pm \frac{5}{4} \cdot \frac{\sqrt{a}}{{\sqrt{a}}^{1 + 1}}$

Add $1$ and $1$ .

$b = \pm \frac{5}{4} \cdot \frac{\sqrt{a}}{{\sqrt{a}}^{2}}$

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

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

$b = \pm \frac{5}{4} \cdot \frac{\sqrt{a}}{{(a^{\frac{1}{2}})}^{2}}$

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

$b = \pm \frac{5}{4} \cdot \frac{\sqrt{a}}{a^{\frac{1}{2} \cdot 2}}$

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

$b = \pm \frac{5}{4} \cdot \frac{\sqrt{a}}{a^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$b = \pm \frac{5}{4} \cdot \frac{\sqrt{a}}{a^{\frac{2}{2}}}$

Rewrite the expression.

$b = \pm \frac{5}{4} \cdot \frac{\sqrt{a}}{a^{1}}$

Simplify.

$b = \pm \frac{5}{4} \cdot \frac{\sqrt{a}}{a}$

Multiply $\frac{5}{4}$ by $\frac{\sqrt{a}}{a}$ .

$b = \pm \frac{5 \sqrt{a}}{4 a}$

Step 4

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$b = \frac{5 \sqrt{a}}{4 a}$

Next, use the negative value of the $\pm$ to find the second solution.

$b = - \frac{5 \sqrt{a}}{4 a}$

The complete solution is the result of both the positive and negative portions of the solution.

$b = \frac{5 \sqrt{a}}{4 a}$

$b = - \frac{5 \sqrt{a}}{4 a}$

$b = \frac{5 \sqrt{a}}{4 a}$

$b = - \frac{5 \sqrt{a}}{4 a}$

Solve for b 16ab^2=25

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