Solve for x ( square root of 7)^(6x)=49^(x-6)

${(\sqrt{7})}^{6 x} = 49^{x - 6}$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{7}$ as $7^{\frac{1}{2}}$ .

${(7^{\frac{1}{2}})}^{6 x} = 49^{x - 6}$

Step 2

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

$7^{\frac{1}{2} (6 x)} = 49^{x - 6}$

Step 3

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

$7^{\frac{1}{2} (6 x)} = 7^{2 (x - 6)}$

Step 4

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

$\frac{1}{2} (6 x) = 2 (x - 6)$

Step 5

Solve for $x$ .

Simplify $\frac{1}{2} \cdot (6 x)$ .

Rewrite.

$0 + 0 + \frac{1}{2} \cdot (6 x) = 2 (x - 6)$

Simplify by adding zeros.

$\frac{1}{2} \cdot (6 x) = 2 (x - 6)$

Cancel the common factor of $2$ .

Factor $2$ out of $6 x$ .

$\frac{1}{2} \cdot (2 (3 x)) = 2 (x - 6)$

Cancel the common factor.

$\frac{1}{2} \cdot (2 (3 x)) = 2 (x - 6)$

Rewrite the expression.

$3 x = 2 (x - 6)$

Simplify $2 (x - 6)$ .

Apply the distributive property.

$3 x = 2 x + 2 \cdot - 6$

Multiply $2$ by $- 6$ .

$3 x = 2 x - 12$

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

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

$3 x - 2 x = - 12$

Subtract $2 x$ from $3 x$ .

$x = - 12$