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
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.
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.
NO! Not "multiply both sides by e". Take "e to the power of both sides" or "take the exponential of both sides."-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.
What happened to the "t"? exp(-.2t+ ln(8))= exp(-.2t)(8)20-y = 8e^(-.2)
y= 20 - 8e^-.2