Solve x in logarithm equation: $2\log_{10} (x-4) - \log_{10}4(x-1) = 0$

Question

I am trying to solve $x$ for $2\log_{10} (x-4) - \log_{10}4(x-1) = 0$

I have the key with the answer 10 and have confirmed this is correct using Wolfram Alpha but which steps should I take to reach that answer?

Answer 1

Start with original problem: $$2 \log_{10} (x-4) - \log_{10}4(x-1) = 0$$

A rule of logarithm: $2 \log_{10} (x-4)=\log_{10}(x-4)^2$:

$$ \log_{10} (x-4)^2 - \log_{10} (4x-4)= 0$$

Another rule of logarithm: Express as the left side of the equation as a single product:

$$ \log_{10} \frac{(x-4)^2}{4x-4}= 0$$

Convert to exponential form:

$$ 10^{\log_{10} \frac{(x-4)^2}{4x-4}}= 10^0$$

or

$$\frac{(x-4)^2}{4x-4}=1$$

and when algebraically rearranging

$$x^2-8x+16=4x-4$$

or

$$x^2-12x+20=0$$

which is now a quadratic equation that you can solve.

Answer 2

Use the properties of logarithms

$$k\log_ax=\log_a(x^k)$$

$$\log_ax-\log_ay=\log_a\frac xy$$

$$\log_ax=0\iff x=1$$