Question

Using the change of base formula, how do you evaluate `log_3 12` (round to the nearest hundredth)?

Answer 1

`color(blue)(2.26)`

Explanation:

Change of base:

Let `color(white)(88)y=log_b(a)`

`y=log_b(a)=> b^y=a`

Taking logs of `b^y=a` with a different base log:

`ylog_c(b)=log_c(a)`

`y= (log_c(a))/log_c(b)`

But `y= log_b(a)`

`:.`

`log_b(a)=(log_c(a))/log_c(b)`

So to change base we take the log of the number divided by the log of the base.

Using base 10 logs:

`log_3(12)=log_10(12)/log_10(3)~~2.261859`

To nearest hundredth: `color(blue)(color(white)(88)2.26)`

Answer 2

`log_3 12=2.262`

Explanation:

According to the change of base formula `log_b a=loga/logb`

Hence `log_3 12`

= `log12/log3`

= `1.0792/0.4771`

= `2.262`