# Math Help - Solving for an exponential variable....did I do this right?

1. ## Solving for an exponential variable....did I do this right?

N= N0/dn
First, I multiplied both sides by dn : dn = N0/N
Then I took the log of both sides: n log d = log(N0/N)
Finally, I divided both sides by log d: n = log(N0/N)/log d
Does this look legit?
I was thinking of using the division log rule to manipulate it further: n = log(N0/N)-log d
Is this correct?

Thanks,
~CC

2. This use of the rule is incorrect,

$\displaystyle \ln a -\ln b = \ln \left(\frac{a}{b}\right)$

3. Originally Posted by calvin_coolidge
N= N0/dn
First, I multiplied both sides by dn : dn = N0/N
Then I took the log of both sides: n log d = log(N0/N)
Finally, I divided both sides by log d: n = log(N0/N)/log d
Does this look legit?
I was thinking of using the division log rule to manipulate it further: n = log(N0/N)-log d
Is this correct?

Thanks,
~CC
Up to the point where you say
$n = log(N_0/N) / log d$

I'd say everything looks right. But where is the exponential you mentioned in the title?? The solution for n is
$n = \frac{N_0}{dN}$

Why use logs?

-Dan

4. Sorry, I typed the original post using Word with super and subscripts and copied/pasted it to the page....apparently the super/subscripts didn't transfer over. N0 is N subzero...

So, the original expression should've read:

N = N0/d^n

1. I first isolated d^n to give: d^n = N0/N
2. Then I took the log of both sides: n*log d = log (N0/N)
3. And then divided both sides by log d: n = log(N0/N)/log d

Thanks,

~CC

5. So is the correct use of the rule this:

n*log d = (log N0)/(log N)....?????....

6. Originally Posted by calvin_coolidge
Sorry, I typed the original post using Word with super and subscripts and copied/pasted it to the page....apparently the super/subscripts didn't transfer over. N0 is N subzero...

So, the original expression should've read:

N = N0/d^n

1. I first isolated d^n to give: d^n = N0/N
2. Then I took the log of both sides: n*log d = log (N0/N)
3. And then divided both sides by log d: n = log(N0/N)/log d

Thanks,

~CC
Correct.

You may also write it as n = [log(N0) - log(N)]/log(d)