# Help with logarithms

• May 7th 2011, 02:17 PM
gobblewobble123
Help with logarithms
So, you probably know that the formula for exponential growth is $y=a(1+r)^t$, with $a$ being the original amount, $r$ being the percent of growth, and $t$ being the amount of time. I would like to learn to get $t$ using logarithms.

Hopefully you may be able to help. Thanks in advance!

Gobblewobble123
• May 7th 2011, 02:20 PM
pickslides
Start by taking the log of both sides, any base will do.
• May 8th 2011, 08:00 PM
gobblewobble123
Maybe you could do a step by step example of how to solve for $x$ using logs? I am trying to teach myself and I love examples! I'm not sure exactly what to do here, so anything helps.

If I take the logs of both sides, would it look like this: $log_10 {a(1+r)}=log_10 {y}$?

Gobblewobble123
• May 8th 2011, 08:02 PM
gobblewobble123
Maybe you could do a step by step example on how to solve the problem? Examples are awesome!

If I took the log of both sides, would it look like this: log_10 {a(1+r)}=log_10 {y}?
• May 8th 2011, 08:03 PM
gobblewobble123
Maybe you could do a step by step example on how to solve the problem? Examples are awesome!

If I took the log of both sides, would it look like this: \log_10 {a(1+r)}=\log_10 {y}?
• May 8th 2011, 08:13 PM
TheChaz
Quote:

Originally Posted by gobblewobble123
So, you probably know that the formula for exponential growth is $y=a(1+r)^t$, with $a$ being the original amount, $r$ being the percent of growth, and $t$ being the amount of time. I would like to learn to get $t$ using logarithms.

Hopefully you may be able to help. Thanks in advance!

Gobblewobble123

Since any base will do, I'll pick "e".
ln(y) = ln[a(1 + r)^t] = ln(a) + t*ln(1 + r)
ln(y) - ln(a) = t*ln(1 + r)
ln(y/a) = t*ln(1 + r)
ln(y/a)/ln(1 + r) = t
• May 8th 2011, 11:13 PM
Prove It
Quote:

Originally Posted by TheChaz
Since any base will do, I'll pick "e".
ln(y) = ln[a(1 + r)^t] = ln(a) + t*ln(1 + r)
ln(y) - ln(a) = t*ln(1 + r)
ln(y/a) = t*ln(1 + r)
ln(y/a)/ln(1 + r) = t

It's easier if you divide both sides by a first, then take the logarithms, but it doesn't really make any difference.
• May 8th 2011, 11:38 PM
TheChaz
Quote:

Originally Posted by Prove It
It's easier if you divide both sides by a first, then take the logarithms, but it doesn't really make any difference.

Funny - after having demonstrated this a few times to students today, I was so distracted by the notational difference (we use A = P(1 + r/n)^{nt}) that I skipped the first step!

While the result is the same, I like to approach problems with invertible operations by composing their inverses (in reverse order), so dividing by a should have been first :)