How do you evaluate #log_sqrt7 49#?

Question

1952 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-04T11:33:37

#4#

We should attempt to write #49# as a power of #sqrt7#.

First, simply rewrite #49# as #7^2#.

#log_sqrt7 49=log_sqrt7(7^2)#

This can be rewritten using the rule:

#log_a(b^c)=c*log_ab#

Thus, we have

#log_sqrt7(7^2)=2log_sqrt7 7#

Now, we can write that #7=(sqrt7)^2#.

#2log_sqrt7(sqrt7)^2#

Rewrite using the rule we found earlier.

#2*2log_sqrt7sqrt7#

Note that

#log_aa=1#

So we are just left with

#2*2log_sqrt7sqrt7=4#

We can test this by using our rules of logarithms. We said that:

#log_sqrt7 49=4#

This means that #(sqrt7)^4=49#

#sqrt7 xx sqrt7xxsqrt7xxsqrt7=7xx7=49#