Question

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

`-aln(a-x)=ln(t-b)`
Solve for x.

Answer 1

`x = a - root(a)(1/(t-b))" "` if `a != 0`

`x < 0" "` if `a = 0`

Explanation:

Your equation looks like this

`-a * ln(a-x) = ln(t-b)`

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

`a -x > 0" "` and `" " t -b > 0" " " "color(orange)("(*)")`

since you can't take the natural log of zero or a negative number while working in `RR`.

Next, we need to split into cases `a = 0` and `a != 0`

Case `bb(a = 0)`

In this case we have

`0 = ln(t-b)`

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 `bb(a != 0)`

With `a != 0` we can first divide both sides of the equation by `-a` to get

`ln(a - x) = ln(t-b) * -1/a`

Then taking exponents of both sides

`e^(ln(a-x)) = e^(ln(t-b)*-1/a)`

Note that if `r > 0` then

`e^ln(r) = r`

Also if `A > 0` and `B, C in RR`, then

`A^(BC) = (A^B)^C`

So our equation simplifies to

`a-x = (t-b)^(-1/a)`

Hence

`x = a - (t-b)^(-1/a) = a - root(a)(1/(t-b))`

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

`1/(t-b) > 0`

and the principal `a`th root represented by

`root(a)(1/(t-b))`

is a well-defined positive Real number.