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.

This is a bit hard to read. Do you mean $2\log_2(x-6)-\log_2(x)=3$? — lulu, Jan 22 '19 at 17:52
@lulu Yes, I mean that. Sorry, I'm not sure how to format it correctly — Grimestock, Jan 22 '19 at 17:53

score 2 · Answer 1 · edited Jan 22 '19 at 17:58

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.

edited Jan 22 '19 at 17:58

Boris Valderrama

768
6
19

answered Jan 22 '19 at 17:52

Klaus

10,578

score 2 · Answer 2 · answered Jan 22 '19 at 17:56

2

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

answered Jan 22 '19 at 17:56

Dr. Sonnhard Graubner

95,283

score 2 · Accepted Answer · answered Jan 22 '19 at 17:57

2

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.

answered Jan 22 '19 at 17:57

kcborys

574

With this would I get x^(3 - 6^6)/x^3? – Grimestock Jan 22 '19 at 18:05
1

Raise $2$ to the power of each side. You should get $\frac{(x-6)^2}{x} = 8$. – kcborys Jan 22 '19 at 18:06
So (x^2-36)/x=8 to x-36=8? – Grimestock Jan 22 '19 at 18:09
1

$(x-6)^2 = (x-6)(x-6) \not= x^2 - 36$ – kcborys Jan 22 '19 at 18:10
Thank you! So (x-6)(x-6) / x = (x^2 -12x+36) /x. But how would I divided 36 by x? I would multiply, right? – Grimestock Jan 22 '19 at 18:13
1

Correct. Then you will have $x^2 - 12x + 36 = 8x$. – kcborys Jan 22 '19 at 18:16
So I would get (x-18)(x-2) to get x=18 and x=2 but then I would have to check to see if there are extraneous roots? – Grimestock Jan 22 '19 at 18:16
Cool. Thanks! The answer would just be x=18 because 2 wouldn't work because we can't log a negative number? – Grimestock Jan 22 '19 at 18:18
1

Correct, because if you plug $x=2$ back in, you have a term of $\log_{2}(-4)$ which is not possible – kcborys Jan 22 '19 at 18:18
Thank you! Sorry that took so long. – Grimestock Jan 22 '19 at 18:19

Boris Valderrama · Answer 4 · 2019-02-03T05:20:12.380

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 $x=2$

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

4 Answers4