anonymous
  • anonymous
log base 2 fifth root of 32 ?!?!?!?!?!??!?!?!?!!?!??! PLEASE HELP
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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Xishem
  • Xishem
\[\log_{2} \sqrt[5]{32}=\log_{2}2=1\]
anonymous
  • anonymous
what?
anonymous
  • anonymous
how did you get log base 2 then 2?

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anonymous
  • anonymous
the fifth root of 32 is 2, because \[2^5=32\]
anonymous
  • anonymous
then shouldn't the answer be five ?
anonymous
  • 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
  • anonymous
\[\log_2(2^5)=5\] yes
anonymous
  • anonymous
but you have \[\log_2(2)\]
anonymous
  • anonymous
wait..what?!
anonymous
  • anonymous
\[2^{y} = 2^{5} \]
anonymous
  • anonymous
ok lets go slow
anonymous
  • anonymous
if your question was \[\log_2(2^5)\] the answer would certainly be 5 correct?
anonymous
  • anonymous
i.e. \[\log_2(32)=5\]
anonymous
  • anonymous
wait one sec let me soak this in
anonymous
  • anonymous
ok
anonymous
  • anonymous
so you're telling me that 2^5 does not equal fifth root of 32?
anonymous
  • anonymous
yes i am. 2^5 = 32 not the fifth root of 32
anonymous
  • anonymous
so what is it? and how would u find what it is ?
anonymous
  • anonymous
i want to know the fifth rote of 32
anonymous
  • anonymous
but since 2^5 = 32 then we know that \[\sqrt[5]{32}=2\]
anonymous
  • anonymous
it is 32^ 1/5
anonymous
  • anonymous
you can think of it that way, yes
anonymous
  • anonymous
im not understanding what you're telling me
anonymous
  • anonymous
which means \[(32)^{\frac{1}{5}}=(2^5)^{\frac{1}{5}}=2^1=2\]
anonymous
  • anonymous
in english: two the the fifth power is 32, therefore the fifth root of 32 is 2
anonymous
  • anonymous
i think i understand that. thank you very much. can you help me with a few more?
anonymous
  • anonymous
post i will reply
anonymous
  • anonymous
\[3 \log _{7} \sqrt[6]{49}\]
anonymous
  • anonymous
ok lets do this the easy way. first of all \[\log(x^n)=n\log(x)\]
anonymous
  • anonymous
yep. i understand tha : - )
anonymous
  • anonymous
and we can rewrite \[\sqrt[6]{49}\] as \[49^{\frac{1}{6}}\] yes?
anonymous
  • anonymous
ummm im kind of confused on those
anonymous
  • anonymous
is that just a general rule ?
anonymous
  • anonymous
you need to understand exponential notation for logs, because logs are exponents.
anonymous
  • anonymous
yes for example \[\sqrt{x}=x^{\frac{1}{2}}\] \[\sqrt[3]{x}=x^{\frac{1}{3}}\] etc
anonymous
  • anonymous
and in general \[\sqrt[n]{x}=x^{\frac{1}{n}}\]
anonymous
  • anonymous
okay awesome. just wrote that on my rules sheet XD
anonymous
  • 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
  • anonymous
\[\frac{1}{6}\times 3\log_7(49)\] \[\frac{1}{2}\log_7(49)\] and idea what to do next?
anonymous
  • anonymous
i really have no idea :(
anonymous
  • anonymous
how can you write 49 with an exponent?
anonymous
  • anonymous
what would u do if you have a number in front of the log? like the 1/2
anonymous
  • anonymous
hold on, lets just take care of the 49 first. we can do this problem a different way second
anonymous
  • anonymous
we would write \[49=7^2\]
anonymous
  • anonymous
and then you have \[\frac{1}{2}\log_77^2)\] which should be easy now
anonymous
  • anonymous
how would we incorporate the 1/2?
anonymous
  • anonymous
since \[\log_b(b^n)=n\]
anonymous
  • anonymous
but its 1/2 log
anonymous
  • anonymous
we leave it for a moment. the one half out front is just a number, let it sit there
anonymous
  • anonymous
so the answer is 2
anonymous
  • anonymous
we get \[\frac{1}{2}\log_7(7^2)=\frac{1}{2}\times 2\]
anonymous
  • anonymous
the answer to \[\log_7(7^2)=2\] and then \[\frac{1}{2}\times 2=1\]
anonymous
  • anonymous
oh okay!!!!! wow i get that!
anonymous
  • anonymous
if you like we can do this problem a different way, but maybe one way is enough, you tell me
anonymous
  • anonymous
answer is 1
anonymous
  • anonymous
yes it is 1
anonymous
  • anonymous
i like this way. do u mind helping me with more?
anonymous
  • anonymous
go ahead
anonymous
  • anonymous
\[50 \log 5 \sqrt{125} \]
anonymous
  • anonymous
so we would have
anonymous
  • anonymous
5 ^ x = square route 125 ^ 50?
anonymous
  • anonymous
oh yikes leave the 50 out front!
anonymous
  • 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
  • anonymous
okay. lol so i have 50 log base 5^x = 5^ 3
anonymous
  • anonymous
i believe that is wrong
anonymous
  • anonymous
close, you just forget the square root
anonymous
  • anonymous
but i don't have the square route anymore i got it al out of the square rote
anonymous
  • anonymous
\[125=5^3\] and so \[\sqrt{125}=5^{\frac{3}{2}}\]
anonymous
  • anonymous
i dont understand that
anonymous
  • anonymous
just a moment ago we found out that \[\sqrt{x}=x^{\frac{1}{2}}\] right?
anonymous
  • anonymous
\[\sqrt{125}=125^{\frac{1}{2}}=(5^3)^{\frac{1}{2}}=5^{\frac{3}{2}}\]
anonymous
  • anonymous
yes. oh okay i understand
anonymous
  • anonymous
so then what would we have
anonymous
  • anonymous
so now we have \[50\log_5(5^{\frac{3}{2}})\] \[50\times \frac{3}{2}\] etc
anonymous
  • anonymous
as soon as you say \[\log_b(\text{whatever})\] you should think of ways to write "whatever" as a power of b
anonymous
  • anonymous
*see
anonymous
  • anonymous
oh okay!!! wow!
anonymous
  • anonymous
can u give me an example to try?
anonymous
  • anonymous
\[\log_3(\sqrt[5]{27})\]
anonymous
  • anonymous
can u do one with a number in front of the log
anonymous
  • 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
  • anonymous
okay let me get my paper lol
anonymous
  • anonymous
6?
anonymous
  • anonymous
yup!
anonymous
  • anonymous
OMG
anonymous
  • anonymous
really? i got it right?
anonymous
  • anonymous
don't be so shocked, you will get the hang of it
anonymous
  • anonymous
can u help me with more? lol
anonymous
  • anonymous
after a couple practice problems you will think "not much to this"
anonymous
  • anonymous
ok we can do another before i go
anonymous
  • anonymous
\[\log _{3} 4 - 1/2 \log _{3} (6x-5)\]
anonymous
  • anonymous
holy moly what are the instruction, write as a single log?
anonymous
  • anonymous
lol i have no idea what to do at all
anonymous
  • anonymous
first you need to make sure what the instructions are. what does it say before the problem?
anonymous
  • anonymous
all it says is : solve
anonymous
  • anonymous
maybe u an help me with a different one instead?
anonymous
  • anonymous
you cannot "solve' because there is not an equation here. maybe it says "write as one log"
anonymous
  • anonymous
which we can do if you like
anonymous
  • anonymous
sure lets write it as one log
anonymous
  • 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
  • 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
  • anonymous
that is the answer?
anonymous
  • anonymous
yes, although the text did not seem to provide a question
anonymous
  • 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
  • anonymous
you too, good luck

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