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

Question

3592 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-11T12:27:46

$color(blue)(2.26)$

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 · 2017-11-11T12:28:38

$log_3 12=2.262$

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

Hence $log_3 12$

= $log12/log3$

= $1.0792/0.4771$

= $2.262$

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