# Logs

• Sep 8th 2011, 06:23 PM
crazycarlos61
Logs
Need some help to tell me what I'm doing wrong.

solve for y:

2e^(-y/3) = t + 4
ln(2e^(-y/3)) = ln(t+4)
-y/3(2) = ln(t+4)
y= -(3/2)ln(t+4)

ln(20-y) = -.2t + ln8
*Mult both sides by e
20-y = 8e^(-.2)
y= 20 - 8e^-.2
• Sep 8th 2011, 06:32 PM
TKHunny
Re: Logs
One thing at a time...

$2\cdot e^{-y/3} = t+4$

Try simple things first. Divide by 2

$e^{-y/3} = (t+4)/2$

Now the logarithm.

$-y/3 = ln\left(\frac{t+4}{2}\right)$

Can you finish?

On the second, think really, REALLY hard about "multiply by e" to remove a logarithm. Until you see that this is very, VERY wrong, keep thinking about it.
• Sep 8th 2011, 06:45 PM
crazycarlos61
Re: Logs
I'll continue on the 2nd but for the first I just got y = -3ln((t+4)/2) Which I still think is wrong.
• Sep 8th 2011, 08:26 PM
TKHunny
Re: Logs
No, that's good, but it can be written a few other ways. Figure out how to get this one:

$y = 3\cdot ln\left(\frac{2}{t+4}\right)$

You should also worry about the Domain.
• Sep 9th 2011, 05:13 AM
HallsofIvy
Re: Logs
Quote:

Originally Posted by crazycarlos61
Need some help to tell me what I'm doing wrong.

solve for y:

2e^(-y/3) = t + 4
ln(2e^(-y/3)) = ln(t+4)

Here is your error: ln(ab)= ln(a)+ ln(b), NOT a ln(b). ln (2e^{-y/3})= ln(2)+ ln(e^{-y/3})= ln(2)- y/3, not "-y/3(2)". To solve for y, you would then subtract ln(2) from both sides which would give you a factor of 2 in the denominator inside the logarithm. That is, this is the same as first dividing both side by 2 as TKHunny suggested.

Quote:

-y/3(2) = ln(t+4)
y= -(3/2)ln(t+4)

ln(20-y) = -.2t + ln8
*Mult both sides by e
NO! Not "multiply both sides by e". Take "e to the power of both sides" or "take the exponential of both sides."
What you did is correct but thinking "multiply" when you don't mean that will lead you into errors eventually.

Quote:

20-y = 8e^(-.2)
y= 20 - 8e^-.2
What happened to the "t"? exp(-.2t+ ln(8))= exp(-.2t)(8)