Solve for x log base 6 of (x-4)^2=2

$\log_{6} ({(x - 4)}^{2}) = 2$

Step 1

Write in exponential form.

For logarithmic equations, $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ such that $x > 0$ , $b > 0$ , and $b \neq 1$ . In this case, $b = 6$ , $x = {(x - 4)}^{2}$ , and $y = 2$ .

$b = 6$

$x = {(x - 4)}^{2}$

$y = 2$

Substitute the values of $b$ , $x$ , and $y$ into the equation $b^{y} = x$ .

$6^{2} = {(x - 4)}^{2}$

Step 2

Solve for $x$ .

Rewrite the equation as ${(x - 4)}^{2} = 6^{2}$ .

${(x - 4)}^{2} = 6^{2}$

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

$| x - 4 | = | 6 |$

Solve for $x$ .

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$x - 4 = \pm | 6 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $6$ is $6$ .

$x - 4 = \pm 6$

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.

$x - 4 = 6$

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

Add $4$ to both sides of the equation.

$x = 6 + 4$

Add $6$ and $4$ .

$x = 10$

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

$x - 4 = - 6$

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

Add $4$ to both sides of the equation.

$x = - 6 + 4$

Add $- 6$ and $4$ .

$x = - 2$

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

$x = 10, - 2$

Solve for x log base 6 of (x-4)^2=2

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