Confused by logarithmic properties

Question

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!

Answer 1

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:

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

Answer 2

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

Answer 3

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