Solving logarithmic equation $2\log _{2}(x-6)-\log _{2}(x)=3$

Question

This is the question: $$2\log_{2} (x-6)-\log_{2} (x)=3$$

I think I would combine the two on the left to make $2\log_{2}\big({x-6\over x}\big) = 3$ but I'm stuck at what to do with the $2$ in front of the log. Would I divide it out to get $\log_{2}\big({x-6\over x}\big) = \tfrac{3}{2}$ or change to equation to exponential form?

Any help would be greatly appreciated as I've been stuck on this question for a while.

Answer 1

You can't quite use your first step. First you should convert $2\log_{2}(x-6)$ to $\log_{2}(x-6)^2$, and then you can apply the subtraction of logs property to get $\log_{2}\frac{(x-6)^2}{x} = 3$. Then exponentiate to get rid of the logs and you should soon find yourself with a quadratic equation that you should be able to solve.

Answer 2

Use the rules $\log\big(\frac{a}{b}\big) = \log(a) - \log(b)$ and $\log(a^n) = n\log(a)$ to write everything as one logarithm. Then exponentiate.

Answer 3

Hint: You can write $$\log_{2}{(x-6)^2}-\log_{2}{6}=3$$ and by the hint above $$\log_{2}\frac{(x-6)^2}{x}=3$$

Answer 4

With the basics rules: $$ \log_a(b)=x \iff a^x=b \label{1}\tag{1}$$ $$ \log_a(b^n)=n\log_a(b) \label{2}\tag{2}$$ $$ \log_a(b) + \log_a(c)=\log_a(bc) \label{3}\tag{3}$$ You can solve this equation: $$ 2\log_{2} (x-6) - \log_2(x)=3$$ from \eqref{2}: $$ \log_{2} \big[(x-6)^2\big] - \log_2(x)=3$$ from \eqref{2}: $$ \log_{2} \big[(x-6)^2\big] + \log_2\big(x^{-1}\big)=3$$ $$ \log_{2} \big[(x-6)^2\big] + \log_2\Big(\frac{1}{x}\Big)=3$$ from \eqref{3}: $$ \log_{2} \Bigg[\frac{(x-6)^2}{x}\Bigg]=3$$ from \eqref{1}: $$ 2^3 = \frac{(x-6)^2}{x} $$ $$ 8 = \frac{(x-6)^2}{x} $$ $$ 8x = (x-6)^2 $$ $$ 8x = x^2-12x+36 $$ $$ 0 = x^2-20x+36 $$ Here, $-20 = -18-2$ and $36=(-18)\cdot(-2)$, then: $$ 0 = (x-18)(x-2) $$ Thus $x=18$ or $x=2$