I am not a regular math user. I need to solve an equation to apply on timeseries variables.
I have value of $A$ but don't have value of $B$ variable. Here, time series variables (mentioned as variable $A$) with known values are canopy cover values (range $40$ to $100$). Whereas variable $B$ (unknown values) are the leaf area index (LAI). The output LAI values (variable $B$) might be in $0.5$ to $10$.
The equation:
$$ A = 94\left(1 - e^{-0.43 B}\right)^{0.52}$$
(In MS Excel: A = 94*POWER(1-EXP(-0.43*B),0.52)})
What would be the equation that can be applied to get value of "$B = \text{expression}$"?
This might be quite simple, but for me it's not. Please help.
The equation is easily solved for $B$ in terms of $A$ by "reversing" each of the operations that appears on the right-hand side:
$$ A = 94(1 - e^{-0.43 B})^{0.52} $$
$$ A/94 = (1 - e^{-0.43 B})^{0.52} $$
$$ (A/94)^{1/0.52} = 1 - e^{-0.43 B} $$
$$ e^{-0.43 B} = 1 - (A/94)^{1/0.52} $$
$$ -0.43 B = \ln (1 - (A/94)^{1/0.52}) $$
Therefore:
$$ B = \frac{\ln (1 - (A/94)^{1/0.52})}{-0.43} $$
This derivation of $B$ and especially the last line make it clear that we want $A$ to be nonnegative and less than $94$ in order for there to be a corresponding real value of $B$. Indeed the last line gives $B$ as the ratio of two negative numbers, so that $B\gt 0$ when $0 \lt A \lt 94$. This is mostly consistent with what the Question says about "canopy cover values" except those in the range for $A$ at or above $94$.