# Solve for x. `log_5(x-4)=log_7(x)`

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### 1 Answer

This question may seem a bit tricky because these are logs of different bases, but you can take advantage of the change-of-base formula to solve the problem. Here is how the change-of-base formula works:

`log_a(b) = lnb/lna`

I only use the natural logarithm, base `e`, because all calculators and computers have it as an option. You can also use any other base you want, like base 10, or even base `pi`!

So, let's start by converting these logarithms into natural logarithms:

`ln(x-4)/ln5 = lnx/ln7`

Now, subtract `lnx/ln7` from both sides:

`ln(x-4)/ln5 - lnx/ln7 = 0`

Multiply both sides by `ln5*ln7` to get rid of the denominators:

`ln7ln(x-4) - ln5lnx = 0`

Now, we can use the power rule of logarithms to make exponents out of `ln7` and `ln5`:

`ln(x-4)^ln7 - lnx^ln5 = 0`

Finally, let's use the division rule of logarithms to get all terms inside of a single logarithm:

`ln((x-4)^ln7/x^ln5) = 0`

Now, allow both sides of the equation to be powers of `e`, cancelling out the logarithm on the left:

`(x-4)^ln7/x^ln5 = 1`

Here's the tough part. There is no good way to solve this equation from here, at least that I know of!

Using a computer algebra system confirms the lack of an algebraic solution method to this problem. However, numerically, you can find the following solution:

`x ~~ 11.584`

Not a very satisfying answer, but I hope it helps you out!

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