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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.

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

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

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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.

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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 $x=2$

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