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I distributed the the log $$\log_2 x^2 - \log_2 2x - \log_2 27 = 3$$ but I am stuck at that point. Any hints?

wckronholm
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Cetshwayo
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    you cannot distribute $\log$. Please note $\log(A+B) \neq \log A+\log B$. – Anurag A Jul 01 '14 at 01:21
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    As often happens, many of the posted answers are more complicated than they need to be. I hope my posted answer is as simple as the answer really is. At any rate, you cannot distributed logarithms like that. I've up-voted the question and some of the answers. Why hasn't anyone else up-voted the question? – Michael Hardy Jul 01 '14 at 17:40

5 Answers5

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Just use definition. $\log_b A=c$ means $b^c=A$. Thus your question becomes. $$x^2-5x-28=2^3$$ Now solve for $x$.

Anurag A
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    This is really the critical point to understand. Somehow it seems teachers don't always impress upon students that a logarithm is just the exponent required to turn the base ($b$) into the given value ($A$). If the student doesn't understand this, there is no hope of understanding how to work with logarithms. +1. – MPW Jul 01 '14 at 03:19
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The first thing is that you can't distribute the $\log$, since it isn't a number, it's a function that doesn't obey linearity, i.e. $\log(\alpha+\beta)\neq\log\alpha+\log\beta$.

You can solve this problem this way: (move your mouse above each step to know which law was used) $$\eqalign{\require{action}\log_2(x^2-5x-28)=3 \ \ &\Rightarrow \ \ \mathtip{2^{\log_2(x^2-5x-28)}=2^3}{\displaystyle\rm because\, a=b \iff c^a=c^b. } \\ &\Rightarrow \ \ \mathtip{x^2-5x-28=8}{\rm \displaystyle\rm because\, b^{\displaystyle\log_bc}=c. } \\ &\Rightarrow \ \ \mathtip{x^2-5x-36=0}{\text{adding $-8$ to both sides}}.}$$ The last equation can be solved using the quadratic formula. And after you get your solutions, check them.

Hakim
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  • Thanks for reminding me. So when I use the laws of logarithm I should look at it them as tools to solve equations not necessarily part of them? – Cetshwayo Jul 01 '14 at 01:26
  • @Utvecklaochförenkla What do you mean by not necessarily part of them? – Hakim Jul 01 '14 at 01:31
  • Well I have done about 10 in the past 20 minutes. I guess I mean that they are tools to solve equations where one is seeking the base of an exponent or the exponent itself. Am I correct? – Cetshwayo Jul 01 '14 at 02:04
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    @Utvecklaochförenkla The laws of logarithms can be considered as tools to solve equations. – Hakim Jul 01 '14 at 02:20
  • It would be -28-8 in your quadratic equation! – MonK Jul 01 '14 at 17:34
  • @Sid Thanks, fixed! :-) – Hakim Jul 01 '14 at 18:43
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Rather than distribute the log, exponentiate both sides:

$$2^{\left(\log_2(x^2 − 5x − 28)\right)} = 2^3$$

wckronholm
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  • How did you see this? Was it because the 3 was isolated? – Cetshwayo Jul 01 '14 at 01:23
  • No. It's more that this way the $2$ and the $\log_2$ cancel. This is really the definition of the logarithm. Since there is just one isolated log term, this method eliminates the log and leave a quadratic. – wckronholm Jul 01 '14 at 01:25
  • As a side note, once you do find the solutions to the quadratic, don't forget to check that they work. Sometimes this approach leads to erroneous solutions. – wckronholm Jul 01 '14 at 01:26
  • @Utvecklaochförenkla : The way you see something like this is to recall that $\log_b a =c$ if and only if $b^c=a$. ${}\qquad{}$ – Michael Hardy Jul 01 '14 at 17:37
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How exactly you distribute the log?

$\log A + \log B= \log AB$ but

$\log(A+B) \ne \log A+\log B$

What you are looking at is a quadratic equation.

$x^2-5x-28=2^3$

Because, if $y=\log_2x$, then $2^y=x$

$x^2-5x-28-8=0$

$x^2-5x-36=0$

$(x-9)(x+4)=0$

Michael Hardy
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MonK
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$$ \log_2\text{something} = 3 \text{ if and only if }2^3=\text{something}. $$

That will get you the solution.

Michael Hardy
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