# Refresh on logarithms

**URL:** <https://boards.straightdope.com/t/refresh-on-logarithms/512425>\
**Category:** Factual Questions\
**Created:** [October 3, 2009, 2:16am UTC](https://boards.straightdope.com/t/refresh-on-logarithms/512425 "2009-10-03T02:16:08Z")\
**Posts on this page:** 6\
**Page:** 1

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**Author:** ![sparky\_1](https://avatars.discourse-cdn.com/v4/letter/s/77aa72/32.png) [@sparky\_1](https://boards.straightdope.com/u/sparky_1)\
**Post date:** [October 3, 2009, 2:16am UTC](https://boards.straightdope.com/t/refresh-on-logarithms/512425/1 "2009-10-03T02:16:08Z")

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I am trying to rework Y = X(1 + k)^n and be able to solve for n.

I used to know how.

Many years ago.

Grrrr.

I come up with log(1 + k) no matter what. My recollection is that I can not reduce this further.

Is this correct?

Please note, this is not for homework help. I can’t believe I am stumped by this.

Any good tutorial sites would be appreciated. I’m gonna lose sleep until I figure this out. Damn you brain!

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**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\
**Post date:** [October 3, 2009, 2:28am UTC](https://boards.straightdope.com/t/refresh-on-logarithms/512425/2 "2009-10-03T02:28:28Z")

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You just need two log properties:

log(x^n) = n\*log(x)  
log(xy) = log(x) + log(y)

So expanding using that we get, for any logarithm basis: log(y) = log(x) + n\*log(1+k)

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**Author:** ![Pseudocode](https://avatars.discourse-cdn.com/v4/letter/p/59ef9b/32.png) [@Pseudocode](https://boards.straightdope.com/u/Pseudocode)\
**Post date:** [October 3, 2009, 2:30am UTC](https://boards.straightdope.com/t/refresh-on-logarithms/512425/3 "2009-10-03T02:30:52Z")

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So (1+k)^n = (k^n) \* k.

Since log\_b x = n is equivalent to b^n = x, we have log\_k (Y/XK) = n.

I’m not sure you can simplify more than that.

Edit: This assumes you mean X(1 + k)^n as written and not [X(1 + k)]^n.

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**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\
**Post date:** [October 3, 2009, 2:48am UTC](https://boards.straightdope.com/t/refresh-on-logarithms/512425/4 "2009-10-03T02:48:01Z")

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> [@Pedro](#):
>
> You just need two log properties:
> 
> log(x^n) = n\*log(x)  
> log(xy) = log(x) + log(y)
> 
> So expanding using that we get, for any logarithm basis: log(y) = log(x) + n\*log(1+k)

And of course solving this linear equation in n gives:

n = [log(y) - log(x)]/log(1+k) = log\_(1+k)(y/x)

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<div class="post-metadata">

**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\
**Post date:** [October 3, 2009, 3:02am UTC](https://boards.straightdope.com/t/refresh-on-logarithms/512425/5 "2009-10-03T03:02:42Z")

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> [@](#):
>
> n = [log(y) - log(x)]/log(1+k) = log\_(1+k)(y/x)

Reading that out loud it says that n is the number that, raised over (1+k), equals y/x. Assuming, like the poster above said, that you meant y = x\*[(1+k)^n].

But this is incorrect:

> [@](#):
>
> So (1+k)^n = (k^n) \* k.

k\*(k^n) = k^(n+1) != (1+k)^n

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<div class="post-metadata">

**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\
**Post date:** [October 3, 2009, 3:27am UTC](https://boards.straightdope.com/t/refresh-on-logarithms/512425/6 "2009-10-03T03:27:24Z")

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Maybe a more direct way to do it would be using the property:

log\_a(a^b) = b

So taking the logarithm base (1+k) to both sides of y/x = (1+k)^n would also yield the same answer. But in general calculators don’t have arbitrary log basis so this property comes handy too:

log\_a(b) = log(b)/log(a) for any log basis on the right side.