
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

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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 $\displaystyle x = \log_ay$, then $\displaystyle a^x = y$.
Can you see it now?
Grandad

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.

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

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"

So it's obviously going to be a long decimal number

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.

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?

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.

Ahhh, so I worked it out and got 12 as an answer, similar to that equation..

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


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