3
$\begingroup$

So I've been watching a Khan Academy video that got me confused and I really want to close my gaps so bear with me. I hope that I can understand why I'm failing to understand this. So here are the steps: $$\log_2 \sqrt \frac{32}{\sqrt8}$$ $$log_2(\frac{32}{\sqrt8})^\frac{1}{2}$$ Here comes my question : How can the exponent (from the second step) be put out of the parentheses? Wouldn't this indicate that = $\frac{1}{2}$ to the power of both $32$ and sroot of $8$, which wouldn't be correct because we only un-rooted $32$, not $8$.

Continuing: $$\frac{1}{2}(log_2(\frac{32}{\sqrt8})$$ $$\frac{1}{2}(log_2{32}-log_2{\sqrt8})$$ Second confusion is at this next step:$$\frac{1}{2} (log_2{32}-\frac{1}{2}log_28)$$ Here I get completely confused because I thought we should either do this instead:$\frac{1}{2} (log_2{32})-\frac{1}{2} (log_28)$ or this:$\frac{1}{2} log_2{32}-\frac{1}{2} log_28$ so any explanation would be a relief.

The lasts steps (which I do understand except the $\frac{1}{4}$ but thats because confusion number 2:

$$Distributing: \frac{1}{2}log_2{32}-\frac{1}{4}log_2{8}$$ $$=\frac{5}{2}-\frac{3}{4}$$

Video for reference: https://www.youtube.com/watch?v=TMmxKZaCqe0 (Last part of the vid, from around minute 8:00 to around 9:58)

Thanks in advance!

$\endgroup$
  • $\begingroup$ If the starting point is $\log_2\sqrt{\frac{32}{\sqrt{8}}}$, doing all that stuff is nonsense: the number you want to compute the logarithm of is $\sqrt[4]{\frac{2^{10}}{2^3}}=2^{7/4}$, so the logarithm in base $2$ is $7/4$. $\endgroup$ – egreg Jan 25 '18 at 23:27
  • 1
    $\begingroup$ Greg, if you don't mind, click on the video I'm sure there is some logic behind it (khan doesn't usually teach nonsense) $\endgroup$ – user472288 Jan 25 '18 at 23:31
  • 2
    $\begingroup$ @egreg How is a student expected to know that $\sqrt{\frac {32}{\sqrt 8}} = 2^{\frac 74}$ without doing the "nonsense". The nonsense is nothing more or less than showing that. $\endgroup$ – fleablood Jan 25 '18 at 23:44
  • $\begingroup$ you haven't "unrooted" anything and $\frac 12$ is a power, not a base and we aren't raising it to anything. Look at it this way. Let $a = \frac {32}{\sqrt a}$. Then $\sqrt a = a^{\frac 12}$ (that's the definition of $x^{\frac 12}$ and it is true for all numbers. Not just $a$.) So $\log_2 (\sqrt{\frac {32}{\sqrt 8}} = \log_2 \sqrt a = \log_2 (a)^{\frac 12} = \log_2 (\frac {32}{\sqrt 8})^{\frac 12}$. That's all they are doing. $\endgroup$ – fleablood Jan 25 '18 at 23:49
  • $\begingroup$ $\log_2 \sqrt 8 = \log_2 8^{\frac 12} = \frac 12 \log_2 8$. So $\frac 12(\log_2 32 - \log_2 \sqrt 8) = \frac 12(\log_2 32 - \log_2 8^{\frac 12}) = \frac 12(\log_2 32 - \frac 12 \log_2 8)$. They haven't distributed anything yet. $\endgroup$ – fleablood Jan 25 '18 at 23:52
4
$\begingroup$

You're getting confused what is what.

$\log_2\sqrt{\frac {32}{\sqrt 8}}$. Okay. Let's let $a =\frac {32}{\sqrt 8}$ so we don't get confused in the first step.

$\log_2\sqrt{\frac {32}{\sqrt 8}} = \log_2 \sqrt{a} =$

$\log_2 (a)^{\frac 12} =$

$\frac 12 \log_2 a$.

Okay, we're done with the first step. Let's bring back $a= \frac {32}{\sqrt{8}}$ so we can continue:

$\frac 12 \log_2 a = \frac 12 \log_2 \frac {32}{\sqrt 8}$.

Let's let $b = 32$ and $c = \sqrt 8$ so we don't get confused on the second step:

$\frac 12 \log_2 \frac {32}{\sqrt 8} = \frac 12 \log_2 \frac bc=$

$\frac 12(\log_2 b - \log_2 c)$

Let's bring $32$ and $\sqrt 8$

$=\frac 12(\log_2 32 - \log_2 \sqrt 8)$

Let's let $d= 32; e= 8$ then (so we don't get confused on the third step)

$=\frac 12(\log_2 d - \log_2 \sqrt e)=$

$\frac 12(\log_2 d - \log_2 e^{\frac 12}) =$

$\frac 12(\log_2 d - \frac 12 \log_2 e)$.

Okay, let's bring in the $32$ and $8$ back:

$= \frac 12(\log_2 32 - \frac 12 \log_2 8)$

Here we can finish:

$= \frac 12(5 - \frac 32) = \frac 74$.

$\endgroup$
  • $\begingroup$ Hey Flea, thanks for your answer. But like I said, I don't quite understand why $\frac 12(\log_2 d - \frac 12 \log_2 e)$ shouldn't be $(\frac 12\log_2 d - \frac 12 \log_2 e)$. Because two steps later, we distribute the $(\frac 12$ right? So why is it that we don't distribute the other $(\frac 12$ that we got from turning the root of $e$ into that $(\frac 12$ just like we did with the root of $d$? $\endgroup$ – user472288 Jan 26 '18 at 22:38
  • $\begingroup$ Because $\log_2 d $ is $\log_2 d$. It is NOT $\log_2 \sqrt{d}$. And $\log_2 \sqrt{e}$ is NOT $\log_2 e$. It is $\log_2 \sqrt e$. $e$ HAS a s square root sign and $d$ does NOT. That is all there is to it!. $\endgroup$ – fleablood Jan 27 '18 at 0:12
  • $\begingroup$ If you had $2(7 + 2*5)$ then when you distribute the $2$ you'd get $2*7$ and you'd get $2(2*5)$ you wouldn't figure "Oh, gee, it already has a $2$ it can't have another" or "gee, you are distributing the outside $2$ so we must distribute the inside $2$ as well, because ... well, because it is there." That $\frac 12$ came for the $\sqrt 8$ being $8^{\frac 13}$. That is a property pertaining to THAT term. It has nothing to do with the term $\log_2 32$ that is an entirely different term. $\endgroup$ – fleablood Jan 27 '18 at 0:17
  • $\begingroup$ "just like we did with the root of d?" When?????? $d$ was never under a root sign. Never, ever, ever, ever. $a$ was under a root sign and we dealt with it then. The second root sign is over the $8$ and it has NOTHING to do with the 32. NOTHING AT ALL. $\endgroup$ – fleablood Jan 27 '18 at 0:19
  • $\begingroup$ Let's do this $\log_2 \sqrt[3]{\frac {32}{\sqrt[5]8}} = \log_2(\frac {32}{\sqrt[5]8})^{\frac 13} = \frac 13( \log_2 \frac {32}{\sqrt[5] 8}) = \frac 13(\log_2 32 - \log_2 \sqrt[5]8) = \frac 13(\log_2 32 - \log_2 8^{\frac 15}) = \frac 13(\log_2 32 - \frac 15 \log_2 8)$. Okay, do you understand why we don't "distribute* the $\frac 15$ to $\log_2 32$? How it has absolutely NOTHING to do with the $\log_2 32$? It's the exact same thing. There is NO difference. $\endgroup$ – fleablood Jan 27 '18 at 0:26
2
$\begingroup$

The video is trying hard to confuse you. ;-)

  • First property: $\log_2\sqrt{x}=\frac{1}{2}\log_2 x$

  • Second property: $\log_2 (a/b)=\log_2 a-\log_2 b$.

Now use $x=32/\sqrt{8}$ and $a=32$, $b=\sqrt{8}$, so you have \begin{align} \log_2\sqrt{\frac{32}{\sqrt{8}}} &=\frac{1}{2}\log_2\frac{32}{\sqrt{8}} && \text{first property} \\[4px] &=\frac{1}{2}(\log_2 32-\log_2\sqrt{8}) && \text{second property} \\[4px] &=\frac{1}{2}\Bigl(\log_2 32-\frac{1}{2}\log_2 8\Bigr) && \text{first property} \\[4px] &=\frac{1}{2}\Bigl(5-\frac{3}{2}\Bigr) \\[4px] &=\frac{1}{2}\frac{7}{2}=\frac{7}{4} \end{align}

More easily: $$ \sqrt{\frac{32}{\sqrt{8}}}= \sqrt{\frac{2^5}{2^{3/2}}}= \sqrt{2^{7/2}}=2^{7/4} $$

$\endgroup$
1
$\begingroup$
  1. He wrote $\log_2 \sqrt{\frac{32}{\sqrt8}}=\log_2 \left(\frac{32}{\sqrt8}\right)^\frac{1}{2}$. He means that the square root is still within the logarithm, i.e. $\log_2 \left(\left(\frac{32}{\sqrt8}\right)^\frac{1}{2}\right)$.

  2. $\log_2{\sqrt8}=\log_2{(8^\frac{1}{2})}=\frac{1}{2}\log_2{8}$, so $\frac{1}{2}(\log_2{32}-\log_2{\sqrt8})=\frac{1}{2} (\log_2{32}-\frac{1}{2}\log_28)=\frac{1}{2} \log_2{32}-\frac{1}{4}\log_28=\frac{5}{2}-\frac{3}{4}$

$\endgroup$
  • $\begingroup$ I don't understand both of these but about the second one, why distribute the coefficient 1/2 of the first log over the other. Thats why I wrote that I thought it should either be $\frac{1}{2} (log_2{32})-\frac{1}{2} (log_28)$ or just:$\frac{1}{2} log_2{32}-\frac{1}{2} log_28$? $\endgroup$ – user472288 Jan 25 '18 at 23:24
  • $\begingroup$ @user472288 $\frac{1}{2} (\log_2{32}-\log_2\sqrt{8})=\frac{1}{2}\log_2{32}-\frac{1}{2}\log_2\sqrt{8}=\frac{1}{2}\log_2{32}-\frac{1}{2}(\frac{1}{2}\log_28)=\frac{1}{2}\log_2{32}-\frac{1}{4}\log_28$. Do you understand this? $\endgroup$ – The Phenotype Jan 25 '18 at 23:34
  • $\begingroup$ They aren't distributing anything yet. You have $\log_2 \sqrt 8$ (not $\log_2 8$). An $\log_2 \sqrt 8 = \log_2 8^{\frac 12} = \frac 12 \log_2 8$. So that "second" $\frac 12$ comes for the $\sqrt{}$ sign over then $8$. And it is "attached" only to that one term. $\endgroup$ – fleablood Jan 26 '18 at 0:15
  • $\begingroup$ Notice in the first term of the four part equation you have $\log_2 \sqrt 8$ but in the second term the $\log_2 \sqrt 8$ is replaced with $\frac 12 \log_2 8$. That $\frac 12$ does NOT come from distributing the outside $\frac 12$ it comes for the $\sqrt{}$ in the $\sqrt{8}$. Because $\log_2 \sqrt 8$ (with the root sign) $ = \frac 12 \log_2 8$ (without the root sign). $\endgroup$ – fleablood Jan 26 '18 at 0:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.