Algebra Examples

${(\sqrt[5]{7})}^{x} = 7^{2 x + 3}$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{7}$ as $7^{\frac{1}{5}}$ .

${(7^{\frac{1}{5}})}^{x} = 7^{2 x + 3}$

Step 2

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

$7^{\frac{1}{5} x} = 7^{2 x + 3}$

Step 3

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$\frac{1}{5} x = 2 x + 3$

Step 4

Solve for $x$ .

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Step 4.1

Simplify $\frac{1}{5} x$ .

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Step 4.1.1

Rewrite.

$0 + 0 + \frac{1}{5} x = 2 x + 3$

Step 4.1.2

Simplify by adding zeros.

$\frac{1}{5} x = 2 x + 3$

Step 4.1.3

Combine $\frac{1}{5}$ and $x$ .

$\frac{x}{5} = 2 x + 3$

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

To write $- 2 x$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$\frac{x}{5} - 2 x \cdot \frac{5}{5} = 3$

Step 4.2.3

Combine $- 2 x$ and $\frac{5}{5}$ .

$\frac{x}{5} + \frac{- 2 x \cdot 5}{5} = 3$

Step 4.2.4

Combine the numerators over the common denominator.

$\frac{x - 2 x \cdot 5}{5} = 3$

Step 4.2.5

Simplify the numerator.

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Step 4.2.5.1

Factor $x$ out of $x - 2 x \cdot 5$ .

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Step 4.2.5.1.1

Raise $x$ to the power of $1$ .

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

Step 4.2.5.1.2

Factor $x$ out of $x^{1}$ .

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

Step 4.2.5.1.3

Factor $x$ out of $- 2 x \cdot 5$ .

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

Step 4.2.5.1.4

Factor $x$ out of $x \cdot 1 + x (- 2 \cdot 5)$ .

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

Step 4.2.5.2

Multiply $- 2$ by $5$ .

$\frac{x (1 - 10)}{5} = 3$

Step 4.2.5.3

Subtract $10$ from $1$ .

$\frac{x \cdot - 9}{5} = 3$

Step 4.2.6

Move $- 9$ to the left of $x$ .

$\frac{- 9 \cdot x}{5} = 3$

Step 4.2.7

Move the negative in front of the fraction.

$- \frac{9 x}{5} = 3$

Step 4.3

Multiply both sides of the equation by $- \frac{5}{9}$ .

$- \frac{5}{9} (- \frac{9 x}{5}) = - \frac{5}{9} \cdot 3$

Step 4.4

Simplify both sides of the equation.

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Step 4.4.1

Simplify the left side.

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Step 4.4.1.1

Simplify $- \frac{5}{9} (- \frac{9 x}{5})$ .

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Step 4.4.1.1.1

Cancel the common factor of $5$ .

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Step 4.4.1.1.1.1

Move the leading negative in $- \frac{5}{9}$ into the numerator.

$\frac{- 5}{9} (- \frac{9 x}{5}) = - \frac{5}{9} \cdot 3$

Step 4.4.1.1.1.2

Move the leading negative in $- \frac{9 x}{5}$ into the numerator.

$\frac{- 5}{9} \cdot \frac{- 9 x}{5} = - \frac{5}{9} \cdot 3$

Step 4.4.1.1.1.3

Factor $5$ out of $- 5$ .

$\frac{5 (- 1)}{9} \cdot \frac{- 9 x}{5} = - \frac{5}{9} \cdot 3$

Step 4.4.1.1.1.4

Cancel the common factor.

$\frac{5 \cdot - 1}{9} \cdot \frac{- 9 x}{5} = - \frac{5}{9} \cdot 3$

Step 4.4.1.1.1.5

Rewrite the expression.

$\frac{- 1}{9} (- 9 x) = - \frac{5}{9} \cdot 3$

Step 4.4.1.1.2

Cancel the common factor of $9$ .

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Step 4.4.1.1.2.1

Factor $9$ out of $- 9 x$ .

$\frac{- 1}{9} (9 (- x)) = - \frac{5}{9} \cdot 3$

Step 4.4.1.1.2.2

Cancel the common factor.

$\frac{- 1}{9} (9 (- x)) = - \frac{5}{9} \cdot 3$

Step 4.4.1.1.2.3

Rewrite the expression.

$- - x = - \frac{5}{9} \cdot 3$

Step 4.4.1.1.3

Multiply.

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Step 4.4.1.1.3.1

Multiply $- 1$ by $- 1$ .

$1 x = - \frac{5}{9} \cdot 3$

Step 4.4.1.1.3.2

Multiply $x$ by $1$ .

$x = - \frac{5}{9} \cdot 3$

Step 4.4.2

Simplify the right side.

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Step 4.4.2.1

Simplify $- \frac{5}{9} \cdot 3$ .

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Step 4.4.2.1.1

Cancel the common factor of $3$ .

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Step 4.4.2.1.1.1

Move the leading negative in $- \frac{5}{9}$ into the numerator.

$x = \frac{- 5}{9} \cdot 3$

Step 4.4.2.1.1.2

Factor $3$ out of $9$ .

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

Step 4.4.2.1.1.3

Cancel the common factor.

$x = \frac{- 5}{3 \cdot 3} \cdot 3$

Step 4.4.2.1.1.4

Rewrite the expression.

$x = \frac{- 5}{3}$

Step 4.4.2.1.2

Move the negative in front of the fraction.

$x = - \frac{5}{3}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{5}{3}$

Decimal Form:

$x = - 1.\overline{6}$

Mixed Number Form:

$x = - 1 \frac{2}{3}$