# How do you solve log_b(2) = .105?

Dec 24, 2015

The trick here would be to convert the log function to exponent form and then solve. Please check the explanation for two approaches to solve the same.

#### Explanation:

${\log}_{b} \left(2\right) = 0.105$

The rule
${\log}_{b} \left(a\right) = k \implies a = {b}^{k}$

Using this rule we get

$2 = {b}^{0.105}$

We can solve this by taking $0.105$ root of $2$

${2}^{\frac{1}{0.105}} = b$

$b = 736.12630909184909714688332138981$ using calculator.

Alternate Method

${\log}_{b} \left(2\right) = 0.105$
Using change of base rule which says ${\log}_{b} \left(a\right) = \log \frac{a}{\log} \left(b\right)$
We get $\log \frac{2}{\log} \left(b\right) = 0.105$
Cross multiplying we get $\log \left(2\right) = 0.105 \cdot \log \left(b\right)$
$\log \frac{2}{0.105} = \log \left(b\right)$
$2.8669523396569637639403704259476 = \log \left(b\right)$
#b = 10^2.8669523396569637639403704259476

$b = 736.12630909184909714688332138989$

We can round it as per requirement or instructions.