log base 2 fifth root of 32 ?!?!?!?!?!??!?!?!?!!?!??! PLEASE HELP

- anonymous

log base 2 fifth root of 32 ?!?!?!?!?!??!?!?!?!!?!??! PLEASE HELP

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- Xishem

\[\log_{2} \sqrt[5]{32}=\log_{2}2=1\]

- anonymous

what?

- anonymous

how did you get log base 2 then 2?

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## More answers

- anonymous

the fifth root of 32 is 2, because
\[2^5=32\]

- anonymous

then shouldn't the answer be five ?

- anonymous

so your job is to find
\[\log_2(2)\] which asks the question, "what power do you raise the number "2" to to get an answer of 2?"
and the answer is clearly 1

- anonymous

\[\log_2(2^5)=5\] yes

- anonymous

but you have
\[\log_2(2)\]

- anonymous

wait..what?!

- anonymous

\[2^{y} = 2^{5} \]

- anonymous

ok lets go slow

- anonymous

if your question was
\[\log_2(2^5)\] the answer would certainly be 5 correct?

- anonymous

i.e.
\[\log_2(32)=5\]

- anonymous

wait one sec let me soak this in

- anonymous

ok

- anonymous

so you're telling me that 2^5 does not equal fifth root of 32?

- anonymous

yes i am. 2^5 = 32 not the fifth root of 32

- anonymous

so what is it? and how would u find what it is ?

- anonymous

i want to know the fifth rote of 32

- anonymous

but since 2^5 = 32 then we know that
\[\sqrt[5]{32}=2\]

- anonymous

it is 32^ 1/5

- anonymous

you can think of it that way, yes

- anonymous

im not understanding what you're telling me

- anonymous

which means
\[(32)^{\frac{1}{5}}=(2^5)^{\frac{1}{5}}=2^1=2\]

- anonymous

in english:
two the the fifth power is 32, therefore the fifth root of 32 is 2

- anonymous

i think i understand that. thank you very much. can you help me with a few more?

- anonymous

post i will reply

- anonymous

\[3 \log _{7} \sqrt[6]{49}\]

- anonymous

ok lets do this the easy way.
first of all
\[\log(x^n)=n\log(x)\]

- anonymous

yep. i understand tha : - )

- anonymous

and we can rewrite
\[\sqrt[6]{49}\] as
\[49^{\frac{1}{6}}\] yes?

- anonymous

ummm im kind of confused on those

- anonymous

is that just a general rule ?

- anonymous

you need to understand exponential notation for logs, because logs are exponents.

- anonymous

yes for example
\[\sqrt{x}=x^{\frac{1}{2}}\]
\[\sqrt[3]{x}=x^{\frac{1}{3}}\] etc

- anonymous

and in general
\[\sqrt[n]{x}=x^{\frac{1}{n}}\]

- anonymous

okay awesome. just wrote that on my rules sheet XD

- anonymous

so we start with
\[3\log_7(49^{\frac{1}{6}})\] and then bring the 1/6 right out front as a multiplier (coefficient)

- anonymous

\[\frac{1}{6}\times 3\log_7(49)\]
\[\frac{1}{2}\log_7(49)\] and idea what to do next?

- anonymous

i really have no idea :(

- anonymous

how can you write 49 with an exponent?

- anonymous

what would u do if you have a number in front of the log? like the 1/2

- anonymous

hold on, lets just take care of the 49 first. we can do this problem a different way second

- anonymous

we would write
\[49=7^2\]

- anonymous

and then you have
\[\frac{1}{2}\log_77^2)\] which should be easy now

- anonymous

how would we incorporate the 1/2?

- anonymous

since
\[\log_b(b^n)=n\]

- anonymous

but its 1/2 log

- anonymous

we leave it for a moment. the one half out front is just a number, let it sit there

- anonymous

so the answer is 2

- anonymous

we get
\[\frac{1}{2}\log_7(7^2)=\frac{1}{2}\times 2\]

- anonymous

the answer to
\[\log_7(7^2)=2\] and then
\[\frac{1}{2}\times 2=1\]

- anonymous

oh okay!!!!! wow i get that!

- anonymous

if you like we can do this problem a different way, but maybe one way is enough, you tell me

- anonymous

answer is 1

- anonymous

yes it is 1

- anonymous

i like this way. do u mind helping me with more?

- anonymous

go ahead

- anonymous

\[50 \log 5 \sqrt{125} \]

- anonymous

so we would have

- anonymous

5 ^ x = square route 125 ^ 50?

- anonymous

oh yikes leave the 50 out front!

- anonymous

i mean you are right, but leave it there to make your life easier. think of how to write 125 as a number raised to a power

- anonymous

okay. lol so i have 50 log base 5^x = 5^ 3

- anonymous

i believe that is wrong

- anonymous

close, you just forget the square root

- anonymous

but i don't have the square route anymore i got it al out of the square rote

- anonymous

\[125=5^3\] and so
\[\sqrt{125}=5^{\frac{3}{2}}\]

- anonymous

i dont understand that

- anonymous

just a moment ago we found out that
\[\sqrt{x}=x^{\frac{1}{2}}\] right?

- anonymous

\[\sqrt{125}=125^{\frac{1}{2}}=(5^3)^{\frac{1}{2}}=5^{\frac{3}{2}}\]

- anonymous

yes. oh okay i understand

- anonymous

so then what would we have

- anonymous

so now we have
\[50\log_5(5^{\frac{3}{2}})\]
\[50\times \frac{3}{2}\] etc

- anonymous

as soon as you say
\[\log_b(\text{whatever})\] you should think of ways to write "whatever" as a power of b

- anonymous

*see

- anonymous

oh okay!!! wow!

- anonymous

can u give me an example to try?

- anonymous

\[\log_3(\sqrt[5]{27})\]

- anonymous

can u do one with a number in front of the log

- anonymous

the number out front does not really make any difference, but if you like i can write
\[10\log_3(\sqrt[5]{27})\]

- anonymous

okay let me get my paper lol

- anonymous

6?

- anonymous

yup!

- anonymous

OMG

- anonymous

really? i got it right?

- anonymous

don't be so shocked, you will get the hang of it

- anonymous

can u help me with more? lol

- anonymous

after a couple practice problems you will think "not much to this"

- anonymous

ok we can do another before i go

- anonymous

\[\log _{3} 4 - 1/2 \log _{3} (6x-5)\]

- anonymous

holy moly what are the instruction, write as a single log?

- anonymous

lol i have no idea what to do at all

- anonymous

first you need to make sure what the instructions are. what does it say before the problem?

- anonymous

all it says is : solve

- anonymous

maybe u an help me with a different one instead?

- anonymous

you cannot "solve' because there is not an equation here. maybe it says "write as one log"

- anonymous

which we can do if you like

- anonymous

sure lets write it as one log

- anonymous

by what we know before
\[n\log(x)=\log(x^n)\] so
\[\frac{1}{2}\log(6x-5)=\log((6x-5)^{\frac{1}{2}})=\log(\sqrt{6x-5})\]

- anonymous

also
\[\log(a)-\log(b)=\log(\frac{a}{b})\] so you can write the whole thing as
\[\log_3(\frac{4}{\sqrt{6x-5}})\]

- anonymous

that is the answer?

- anonymous

yes, although the text did not seem to provide a question

- anonymous

okay. thank you so so so much for all your help! maybe i will pass math because of you! :D thanks again and have a wonderful day!!

- anonymous

you too, good luck

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