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

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- Sep 8th 2011, 05:23 PMcrazycarlos61Logs
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, 05:32 PMTKHunnyRe: Logs
One thing at a time...

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

Try simple things first. Divide by 2

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

Now the logarithm.

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

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, 05:45 PMcrazycarlos61Re: 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, 07:26 PMTKHunnyRe: Logs
No, that's good, but it can be written a few other ways. Figure out how to get this one:

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

You should also worry about the Domain. - Sep 9th 2011, 04:13 AMHallsofIvyRe: Logs
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

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