
Solving for t?
Hey all not sure if this should be in precalc or calc. I'm trying to solve the following equation (Von Bertalanffy) for T.
L(t)=Li(1e^(k(TTo)))
L(t)=Length at age t
Li=Theoretical maximum length
k=coefficient of growth
To=Theoretical age at which growth =zero
I have all my other parameters, I just need an equation where I can enter the numbers and find T. Unfortunately I was never great at math...(Crying)
Thanks for any help...

$\displaystyle L_t=L_i (1e^{k(TT_o)})$
Move the L's first;
$\displaystyle \frac{L_t}{L_i} = 1e^{k(TT_o)}$
Move the 1 now.
$\displaystyle 1\frac{L_t}{L_i} = e^{k(TT_o)}$
Use natural log on both sides;
$\displaystyle ln(1\frac{L_t}{L_i} )= ln(e^{k(TT_o)})$
$\displaystyle ln(1\frac{L_t}{L_i} )= k(TT_o)$
Divide by k...
$\displaystyle \frac{ln(1\frac{L_t}{L_i} )}{k}= TT_o$
Add To;
$\displaystyle \frac{ln(1\frac{L_t}{L_i} )}{k} + T_o= T$
$\displaystyle T = \frac{ln(1\frac{L_t}{L_i} )}{k} + T_o$
Here you are :)

Problem $\displaystyle L(t)= L_i(1e^{k(TT_0)}$
Answer:
$\displaystyle 1\frac {L(t)}{L_i} = e^{k(TT_0)}$
$\displaystyle ln(1\frac {L(t)}{L_i}) = ln(e^{k(TT_0)})$
Therefore,
$\displaystyle ln(1\frac {L(t)}{L_i}) = k(TT_0)$
$\displaystyle T= \frac{ln(1\frac {L(t)}{L_i})}{k} + T_0$
:D

Yall are quick....Thanks!