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

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.

• This is a bit hard to read. Do you mean $2\log_2(x-6)-\log_2(x)=3$? – lulu Jan 22 at 17:52
• @lulu Yes, I mean that. Sorry, I'm not sure how to format it correctly – Grimestock Jan 22 at 17:53
• Then note that $n\log a= \log a^n$. – lulu Jan 22 at 17:55

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.

• With this would I get x^(3 - 6^6)/x^3? – Grimestock Jan 22 at 18:05
• Raise $2$ to the power of each side. You should get $\frac{(x-6)^2}{x} = 8$. – kcborys Jan 22 at 18:06
• $(x-6)^2 = (x-6)(x-6) \not= x^2 - 36$ – kcborys Jan 22 at 18:10
• Correct. Then you will have $x^2 - 12x + 36 = 8x$. – kcborys Jan 22 at 18:16
• Correct, because if you plug $x=2$ back in, you have a term of $\log_{2}(-4)$ which is not possible – kcborys Jan 22 at 18:18

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

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.

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 $$x=2$$