Solve for b log base b of 36=2

$\log_{b} (36) = 2$

Step 1

Rewrite $\log_{b} (36) = 2$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$b^{2} = 36$

Step 2

Solve for $b$ .

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

$b = \pm \sqrt{36}$

Simplify $\pm \sqrt{36}$ .

Rewrite $36$ as $6^{2}$ .

$b = \pm \sqrt{6^{2}}$

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

$b = \pm 6$

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.

$b = 6$

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

$b = - 6$

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

$b = 6, - 6$

Step 3

Exclude the solutions that do not make $\log_{b} (36) = 2$ true.

$b = 6$

Solve for b log base b of 36=2

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