Question

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

Answer 1

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(2) = 0.105`

The rule
`log_b(a) = k => a=b^k`

Using this rule we get

`2 = b^0.105`

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

`2^(1/0.105) = b`

`b =736.12630909184909714688332138981` using calculator.

Alternate Method

`log_b(2) = 0.105`
Using change of base rule which says `log_b(a) = log(a)/log(b)`
We get `log(2)/log(b) = 0.105`
Cross multiplying we get `log(2) = 0.105*log(b)`
`log(2)/0.105 = log(b)`
`2.8669523396569637639403704259476 = log(b)`
#b = 10^2.8669523396569637639403704259476

`b=736.12630909184909714688332138989`

We can round it as per requirement or instructions.