# How do you solve this equation? -aln(a-x)=ln(t-b)

## $- a \ln \left(a - x\right) = \ln \left(t - b\right)$ Solve for x.

Apr 17, 2016

$x = a - \sqrt[a]{\frac{1}{t - b}} \text{ }$ if $a \ne 0$

$x < 0 \text{ }$ if $a = 0$

#### Explanation:

$- a \cdot \ln \left(a - x\right) = \ln \left(t - b\right)$

You're dealing with natural logs, so right from the start you know that you must have

$a - x > 0 \text{ }$ and " " t -b > 0" " " "color(orange)("(*)")

since you can't take the natural log of zero or a negative number while working in $\mathbb{R}$.

Next, we need to split into cases $a = 0$ and $a \ne 0$

Case $\boldsymbol{a = 0}$

In this case we have

$0 = \ln \left(t - b\right)$

So

$t - b = 1$

Adding $b$ to both sides we find

$t = b + 1$

The value of $x$ is unconstrained except that we require

$a - x > 0$

Hence

$x < 0$

Case $\boldsymbol{a \ne 0}$

With $a \ne 0$ we can first divide both sides of the equation by $- a$ to get

$\ln \left(a - x\right) = \ln \left(t - b\right) \cdot - \frac{1}{a}$

Then taking exponents of both sides

${e}^{\ln \left(a - x\right)} = {e}^{\ln \left(t - b\right) \cdot - \frac{1}{a}}$

Note that if $r > 0$ then

${e}^{\ln} \left(r\right) = r$

Also if $A > 0$ and $B , C \in \mathbb{R}$, then

${A}^{B C} = {\left({A}^{B}\right)}^{C}$

So our equation simplifies to

$a - x = {\left(t - b\right)}^{- \frac{1}{a}}$

Hence

$x = a - {\left(t - b\right)}^{- \frac{1}{a}} = a - \sqrt[a]{\frac{1}{t - b}}$

Note that since $t - b > 0$ we also have

$\frac{1}{t - b} > 0$

and the principal $a$th root represented by

$\sqrt[a]{\frac{1}{t - b}}$

is a well-defined positive Real number.