# Solving for t?

• June 10th 2010, 09:53 AM
duckcrazed
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...(Crying)

Thanks for any help...
• June 10th 2010, 10:00 AM
Unknown008
$L_t=L_i (1-e^{-k(T-T_o)})$

Move the L's first;

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

Move the 1 now.

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

Use natural log on both sides;

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

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

Divide by -k...

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

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

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

Here you are :)
• June 10th 2010, 10:03 AM
hungthinh92
Problem $L(t)= L_i(1-e^{-k(T-T_0)}$

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