1. ## Solving for t?

Hey all not sure if this should be in pre-calc or calc. I'm trying to solve the following equation (Von Bertalanffy) for T.

L(t)=Li(1-e^(-k(T-To)))

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...

Thanks for any help...

2. $\displaystyle L_t=L_i (1-e^{-k(T-T_o)})$

Move the L's first;

$\displaystyle \frac{L_t}{L_i} = 1-e^{-k(T-T_o)}$

Move the 1 now.

$\displaystyle 1-\frac{L_t}{L_i} = e^{-k(T-T_o)}$

Use natural log on both sides;

$\displaystyle ln(1-\frac{L_t}{L_i} )= ln(e^{-k(T-T_o)})$

$\displaystyle ln(1-\frac{L_t}{L_i} )= -k(T-T_o)$

Divide by -k...

$\displaystyle \frac{ln(1-\frac{L_t}{L_i} )}{-k}= T-T_o$

$\displaystyle \frac{ln(1-\frac{L_t}{L_i} )}{-k} + T_o= T$

$\displaystyle T = \frac{ln(1-\frac{L_t}{L_i} )}{-k} + T_o$

Here you are

3. Problem $\displaystyle L(t)= L_i(1-e^{-k(T-T_0)}$

$\displaystyle 1-\frac {L(t)}{L_i} = e^{-k(T-T_0)}$
$\displaystyle ln(1-\frac {L(t)}{L_i}) = ln(e^{-k(T-T_0)})$
$\displaystyle ln(1-\frac {L(t)}{L_i}) = -k(T-T_0)$
$\displaystyle T= \frac{ln(1-\frac {L(t)}{L_i})}{-k} + T_0$