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

Nov 1, 2015

$x = 2$

#### Explanation:

We would like to have an expression like

${\log}_{4} \left(a\right) = {\log}_{4} \left(b\right)$, 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 $2 = {\log}_{4} \left(16\right)$.

The equation then rewrites as

${\log}_{4} \left(8 x\right) - {\log}_{4} \left(16\right) = {\log}_{4} \left(x - 1\right)$

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. $\log \left(a\right) - \log \left(b\right) = \log \left(\frac{a}{b}\right)$

So, the equation becomes

${\log}_{4} \left(8 \frac{x}{16}\right) = {\log}_{4} \left(x - 1\right)$

Which is of course

${\log}_{4} \left(\frac{x}{2}\right) = {\log}_{4} \left(x - 1\right)$

Now we are in the desired form: since the logarithm is injective, if ${\log}_{4} \left(a\right) = {\log}_{4} \left(b\right)$, then necessarily $a = b$. In our case,

${\log}_{4} \left(\frac{x}{2}\right) = {\log}_{4} \left(x - 1\right) \iff \frac{x}{2} = x - 1$

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