Isolating Variable L in G = log [ aL (1-e^(-bL) + S e^(-bL)]

[amirshi]

Hi,
I need to isolet the variabl L from the equation, I'll appreciate any help:

G = log [ aL (1-e^(-bL) + S e^(-bL)]

I know all the other variables and they are all nombers

Thanks, Shira

 

[arildno]

This is a transcendental equation; you won't be able to to find a closed form expression of L in terms of the other quantities.

 

[M98Ranger]

To isolate the variable L, we need to get it on one side of the equation by itself. To do this, we can use the properties of logarithms and exponentials to simplify the equation. First, let's start by expanding the logarithm using the product rule:

G = log (aL) + log (1-e^(-bL)) + log (S e^(-bL))

Next, we can use the power rule to bring down the exponent in the first term:

G = log (a) + log (L) + log (1-e^(-bL)) + log (S e^(-bL))

Now, we can use the quotient rule to simplify the second and fourth terms:

G = log (a) + log (L) + log [(1-e^(-bL))/e^(-bL)] + log (S)

Using the property of logarithms that states log (a/b) = log (a) - log (b), we can rewrite the third term as:

G = log (a) + log (L) + [log (1) - log (e^(-bL))] + log (S)

Since log (1) = 0, the third term becomes:

G = log (a) + log (L) + [-log (e^(-bL))] + log (S)

Using the property of logarithms that states log (a^b) = b*log (a), we can further simplify the third term:

G = log (a) + log (L) + [bL*log (e)] + log (S)

Since log (e) = 1, the third term becomes:

G = log (a) + log (L) + bL + log (S)

Now, we can use the commutative property of addition to rearrange the terms:

G = log (a) + bL + log (L) + log (S)

Finally, we can combine the second and third terms using the power rule to get:

G = log (a) + bL + log (L*S)

Now, we have isolated the variable L on one side of the equation. To solve for L, we can use the property of logarithms that states log (a*b) = log (a) + log (b). This means that:

log (L*S) = log (a) + bL

Using