Algebra Examples

$49^{2 x^{2} + 3} = 343^{x^{2} + 5}$

Step 1

Create equivalent expressions in the equation that all have equal bases.

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

Step 2

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

$2 (2 x^{2} + 3) = 3 (x^{2} + 5)$

Step 3

Solve for $x$ .

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Step 3.1

Simplify $2 (2 x^{2} + 3)$ .

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Step 3.1.1

Rewrite.

$0 + 0 + 2 (2 x^{2} + 3) = 3 (x^{2} + 5)$

Step 3.1.2

Simplify by adding zeros.

$2 (2 x^{2} + 3) = 3 (x^{2} + 5)$

Step 3.1.3

Apply the distributive property.

$2 (2 x^{2}) + 2 \cdot 3 = 3 (x^{2} + 5)$

Step 3.1.4

Multiply.

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Step 3.1.4.1

Multiply $2$ by $2$ .

$4 x^{2} + 2 \cdot 3 = 3 (x^{2} + 5)$

Step 3.1.4.2

Multiply $2$ by $3$ .

$4 x^{2} + 6 = 3 (x^{2} + 5)$

Step 3.2

Simplify $3 (x^{2} + 5)$ .

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Step 3.2.1

Apply the distributive property.

$4 x^{2} + 6 = 3 x^{2} + 3 \cdot 5$

Step 3.2.2

Multiply $3$ by $5$ .

$4 x^{2} + 6 = 3 x^{2} + 15$

Step 3.3

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

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Step 3.3.1

Subtract $3 x^{2}$ from both sides of the equation.

$4 x^{2} + 6 - 3 x^{2} = 15$

Step 3.3.2

Subtract $3 x^{2}$ from $4 x^{2}$ .

$x^{2} + 6 = 15$

Step 3.4

Move all terms not containing $x$ to the right side of the equation.

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Step 3.4.1

Subtract $6$ from both sides of the equation.

$x^{2} = 15 - 6$

Step 3.4.2

Subtract $6$ from $15$ .

$x^{2} = 9$

Step 3.5

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

$x = \pm \sqrt{9}$

Step 3.6

Simplify $\pm \sqrt{9}$ .

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Step 3.6.1

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{3^{2}}$

Step 3.6.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 3$

Step 3.7

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

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Step 3.7.1

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

$x = 3$

Step 3.7.2

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

$x = - 3$

Step 3.7.3

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

$x = 3, - 3$