How do you express #log 36# in terms of #log 2# and #log 3#?

Question

28716 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-29T09:40:47

#log36=2*(log2+log3)#

To write this expression in terms of #log2# and #log3# you first have to write number #36# as a product of powers of #2# and #3#.

#36=6^2=(2*3)^2#

So you can write:

#log36=log(2*3)^2#

Now you can use the properties of logarythms which say rhat:

After using rule #1# you get:

#log(2*3)^2=2*log(2*3)#

Now if you use rule #2# you get the final result:

#2*log(2*3)=2*(log2+log3)#

Now you can write the final answer:

#log36=2*(log2+log3)#