# Log question

• Jan 8th 2009, 08:11 AM
Slipery
Log question
Hey guys.. I am doing some correspondence work and I was wondering if I could have help with a question. Am I just supposed to do this by trial and error? Seems like there should be an easier way.

http://s180.photobucket.com/albums/x...=logquest2.jpg

Thanks
• Jan 8th 2009, 09:15 AM
Logs
Hello slipery

The definition of a logarithm of a certain number is that power to which the base must be raised to give you the number. So:

If $x = \log_ay$, then $a^x = y$.

Can you see it now?

• Jan 8th 2009, 09:19 AM
Revenantus
Hi Slipery,

You can calculate that value using the following;

y = log7(12)

7^y = 12

y*ln(7) = ln(12)

y = ln(12)/ln(7)

Therefore, log7(12) = ln(12)/ln(7)

This is also known more generally as the 'Change of base formula'.

logA(B) = ln(B)/ln(A)

EDIT: I apologize, I misread the picture you linked - feel free to ignore me.
• Jan 8th 2009, 10:10 AM
Slipery
I don't think you misread my post Revenantus.
And I do understand what you're saying Grandad.

The method you laid out Revantus is the one that I would do myself. I am just wondering whether that is correct.

It is basically saying 7^ log7(12).

First I must figure out log 7(12), and then all I do is work that out as an exponenet for 7. Log 7(12) is what I am having trouble with. I am aware of how to use the scientific calculator to figure the answer out, but none of the examples I am given seem to come out in a decimal like that, so I am just wondering if that is indeed how I am supposed to solve the problem
• Jan 8th 2009, 10:21 AM
Revenantus
Instead of just the mathematical notation, let's think through what each part means.

log7(12) means "The power I must raise 7 to in order to result in 12".

Therefore 7^log7(12) means "I raise 7 to the power that I must raise 7 to in order to result in 12"
• Jan 8th 2009, 10:28 AM
Slipery
So it's obviously going to be a long decimal number
• Jan 8th 2009, 10:32 AM
Revenantus
Not quite. I'll try another example.

e.g. 2^log2(4)

This expression says that I'm raising 2 to the power that it needs to be raised to in order to result in 4.

So, 2^log2(4) = 4

We can verify this easily because log2(4)=2 and 2^2=4.
• Jan 8th 2009, 10:35 AM
Slipery
Yes, but that works out easily because your using 2 and 4... Using 7 and 12 will still result in a decimal, will it not?
• Jan 8th 2009, 10:38 AM
Revenantus
Another approach would be to look back at the two definitions Grandad gave us.

a^x = y and x=loga(y)

By substituting loga(y) in place of x in the first definition we can form;

a^loga(y) = y

Your expression is of the same form as above, try plugging in the values and see what you get.
• Jan 8th 2009, 10:45 AM
Slipery
Ahhh, so I worked it out and got 12 as an answer, similar to that equation..
• Jan 8th 2009, 10:47 AM
Slipery
I understand now.. Its 7 to the 'whatever exponenet it takes 7 to get to 12'.. so obviously my original answer will be 12 as well..

Thanks a lot Revenantus and Grandad
• Jan 8th 2009, 10:48 AM
Revenantus
Excellent, well done.
• Jan 8th 2009, 11:09 AM