# Solve for x: `ln(2x-1) = 4`

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Given the equation:

`ln(2x-1) = 4`

`` First we will rewrite into the exponent form.

" if `ln a = b ==gt a = e^b` "

`==gt (2x-1)= e^4`

`` Now we will solve for x.

`==gt 2x = e^4 +1`

`` `==gt x = (e^4+1)/2 ~~ 27.8`

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To solve this question, we need to remember a certain fact about functions:

`f^-1(f(x)) = x`

This relation means that when we take the inverse of a function, we simply get our original input back out.

This relation is especially relevant for us because (as you can see from the link below) `e^x` is the inverse function of `ln(x)`. In other words,

`e^(ln(x)) = x` and `ln(e^x) = x`

Now, we will use the same principle by setting both sides of the equation as exponents of e:

`e^ln(2x-1) = e^4`

Again, e and ln are inverse functions, so they cancel each other out

`2x - 1 = e^4`

Now, we just solve this like any other equation. First, we'll add 1 to both sides, then we'll divide both sides by 2:

`2x = e^4+1`

`x = (e^4 + 1)/2 = 27.80`

And there you have your x!

There is one danger in solving this problem, though. You have to ensure there are no problems with your domain. In other words, if your x-value produces an undefined function in the original problem, your solution is no good! In our case, if we had an x-value that would force us to take the natural log of a negative number, then that solution clearly cannot be a solution to the problem. In our case, 2(27.80) - 1 is within the domain for ln(x), so we're good.

Therefore, our solution is **x = 27.80**

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