Precalculus Examples

\log_{1.05} (2)

$\log_{1.05} (2)$

Step 1

Rewrite

\log_{1.05} (2)

$\log_{1.05} (2)$ using the change of base formula.

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Step 1.1

The change of base rule can be used if

a

$a$ and

b

$b$ are greater than

0

$0$ and not equal to

1

$1$ , and

x

$x$ is greater than

0

$0$ .

\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}

$\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}$

Step 1.2

Substitute in values for the variables in the change of base formula, using

b = 10

$b = 10$ .

\frac{\log (2)}{\log (1.05)}

$\frac{\log (2)}{\log (1.05)}$

\frac{\log (2)}{\log (1.05)}

$\frac{\log (2)}{\log (1.05)}$

Step 2

Log base

10

$10$ of

1.05

$1.05$ is approximately

0.02118929

$0.02118929$ .

\frac{\log (2)}{0.02118929}

$\frac{\log (2)}{0.02118929}$

Step 3

Log base

10

$10$ of

2

$2$ is approximately

0.30102999

$0.30102999$ .

\frac{0.30102999}{0.02118929}

$\frac{0.30102999}{0.02118929}$

Step 4

Divide

0.30102999

$0.30102999$ by

0.02118929

$0.02118929$ .

14.20669908

$14.20669908$

\log_{1.05} (2)

$\log_{1.05} (2)$

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