How do I solve this logarithmic equation ln

Question

How would I solve an equation like this? I need to find v.

$$a\cdot\ln{(b+v)} = -t + a\cdot\ln{(b+d)} $$

Is it possible to solve something like this using $e$ without excessive complications?

**edited because forgot to actually put v in the equatoin.

Answer 1

$$a\cdot\ln{(b+v)} = -t + a\cdot\ln{(b+d)}$$

Divide by $a\neq 0$ and exponentiate both sides:

$$b+v=e^{-t/a+\ln(b+d)} \implies v = (b+d)e^{-t/a}-b$$

Answer 2

we have $$a(\ln(b+c)-\ln(b+d))=-t$$ and if $$a\ne 0$$ we get $$\ln\left(\frac{b+c}{b+d}\right)=-\frac{t}{a}$$ can you proceed?

Answer 3

First put the natural logs together on one side and factor out the $a$: $$ a \left[\ln (b+v) - \ln(b+d)\right] = -t$$

Now divide both sides by $a$ (we require $a\ne 0$ for this) and use the fact that $\ln X - \ln Y = \ln (X/Y)$ to get: $$ \ln \frac{b+v}{b+d} = -\frac ta$$

Now convert this to an exponential equation. That is, use the fact that $\ln X = Y$ is equivalent to saying $X = e^Y$: $$ \frac{b+v}{b+d} = e^{-t/a}$$

Finally, multiply both sides by $b+d$ and then subtract $b$ from both sides.