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 , then .
Can you see it now?
Jan 8th 2009, 08:19 AM
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, 09:10 AM
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, 09:21 AM
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, 09:28 AM
So it's obviously going to be a long decimal number
Jan 8th 2009, 09:32 AM
Not quite. I'll try another example.
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, 09:35 AM
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, 09:38 AM
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, 09:45 AM
Ahhh, so I worked it out and got 12 as an answer, similar to that equation..
Jan 8th 2009, 09:47 AM
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..