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