What is x if log_4(8x ) - 2 = log_4 (x-1)?

1 Answer
Nov 1, 2015

x=2

Explanation:

We would like to have an expression like

log_4(a)=log_4(b), because if we had it, we could finish easily, observing that the equation would the solved if and only if a=b. So, let's do some manipulations:

  1. First of all, note that 4^2=16, so 2=log_4(16).

The equation then rewrites as

log_4(8x)-log_4(16)=log_4(x-1)

But we're still not happy, because we have the difference of two logarithms in the left member, and we want a unique one. So we use

  1. log(a)-log(b)=log(a/b)

So, the equation becomes

log_4(8x/16)=log_4(x-1)

Which is of course

log_4(x/2)=log_4(x-1)

Now we are in the desired form: since the logarithm is injective, if log_4(a)=log_4(b), then necessarily a=b. In our case,

log_4(x/2)=log_4(x-1) iff x/2 = x-1

Which is easily solve into x=2x-2, which yields x=2