Algebra Examples

\log_{32} (2)

$\log_{32} (2)$

Step 1

Rewrite as an equation.

\log_{32} (2) = x

$\log_{32} (2) = x$

Step 2

Rewrite

\log_{32} (2) = x

$\log_{32} (2) = x$ in exponential form using the definition of a logarithm. If

x

$x$ and

b

$b$ are positive real numbers and

b

$b$ does not equal

1

$1$ , then

\log_{b} (x) = y

$\log_{b} (x) = y$ is equivalent to

b^{y} = x

$b^{y} = x$ .

32^{x} = 2

$32^{x} = 2$

Step 3

Create expressions in the equation that all have equal bases.

{(2^{5})}^{x} = 2^{1}

${(2^{5})}^{x} = 2^{1}$

Step 4

Rewrite

{(2^{5})}^{x}

${(2^{5})}^{x}$ as

2^{5 x}

$2^{5 x}$ .

2^{5 x} = 2^{1}

$2^{5 x} = 2^{1}$

Step 5

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

5 x = 1

$5 x = 1$

Step 6

Solve for

x

$x$ .

x = \frac{1}{5}

$x = \frac{1}{5}$

Step 7

The variable

x

$x$ is equal to

\frac{1}{5}

$\frac{1}{5}$ .

\frac{1}{5}

$\frac{1}{5}$

Step 8

The result can be shown in multiple forms.

Exact Form:

\frac{1}{5}

$\frac{1}{5}$

Decimal Form:

0.2

$0.2$

\log_{32} 2

$\log_{32} 2$

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