Question

simplify ln64-ln8/ln2

Answer 1

I know the first step would be to make it ln64/8
____________
ln2

Answer 2

OK, so it is \(\Large \dfrac{\ln64-\ln8}{\ln2}\)?

Because what you wrote is:
\(\Large \ln64-\dfrac{\ln8}{\ln2}\)

Answer 3

So if it's \(\Large \dfrac{\ln64-\ln8}{\ln2}\) then you are correct, you can start with the rule that the difference of logs is the log of a quotient:

\(\Large \dfrac{\ln64-\ln8}{\ln2}=\dfrac{\ln(64/8)}{\ln2}\)

Now simplify what's inside the parentheses....

Answer 4

Then you'll also be able to use another handy property of logs, usually called the change of base formula (but here, you're using it in the "opposite" of the usual direction):

\(\Large \dfrac{\log_{b}{M}}{\log_{b}{N}}=\log_{N}{M}\)

Answer 5

Once you reach that point, it should all fall into place. :)

Answer 6

thank you very much, i am not sure what to do with with change of base formula, would you be able to walk me through that step?

Answer 7

After you simplified in the fraction, you should have:

\(\Large \dfrac{\ln8}{\ln2}\) right?

Answer 8

yes!

Answer 9

So apply the formula I gave you above:

\(\Large \dfrac{\log_{b}{M}}{\log_{b}{N}}=\log_{N}{M}\) to that expression:

\(\Large \dfrac{\log_{e}{8}}{\log_{e}{2}}=????\)

Notice, M=8, N=2

Answer 10

(see the base e doesn't matter - the change of base formula will allow you to do away with the e. The log expression you end up with is one that you should be able to evaluate easily. :)

Answer 11

okay so therefore log_2(8) is 3, so the answer is 3 - correct?

Answer 12

Exactly! :)

Answer 13

Thank you so much! Great answer !! :)

Answer 14

You're very welcome. :) happy to help!