# Confused by logarithmic properties

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)

• 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 $7/4$. – egreg Jan 25 '18 at 23:27
• 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) – user472288 Jan 25 '18 at 23:31
• @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. – fleablood Jan 25 '18 at 23:44
• 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. – fleablood Jan 25 '18 at 23:49
• $\log_2 \sqrt 8 = \log_2 8^{\frac 12} = \frac 12 \log_2 8$. 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. – fleablood Jan 25 '18 at 23:52

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:

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

• 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 $d$? – user472288 Jan 26 '18 at 22:38
• 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!. – fleablood Jan 27 '18 at 0:12
• 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. – fleablood Jan 27 '18 at 0:17
• "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. – fleablood Jan 27 '18 at 0:19
• 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. – fleablood Jan 27 '18 at 0:26

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

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 $\frac{1}{2}(\log_2{32}-\log_2{\sqrt8})=\frac{1}{2} (\log_2{32}-\frac{1}{2}\log_28)=\frac{1}{2} \log_2{32}-\frac{1}{4}\log_28=\frac{5}{2}-\frac{3}{4}$

• 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:$\frac{1}{2} log_2{32}-\frac{1}{2} log_28$? – user472288 Jan 25 '18 at 23:24
• @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? – The Phenotype Jan 25 '18 at 23:34
• 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. – fleablood Jan 26 '18 at 0:15
• 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). – fleablood Jan 26 '18 at 0:20