anonymous
  • anonymous
Write in exponential form. (Will write in equation editor)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[5^{\log5}^{25}\]
anonymous
  • anonymous
wait..that came out funny..
anonymous
  • anonymous
\[5^{\log5}\]

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anonymous
  • anonymous
There is suppoesd to be a 25 next to log 5
UnkleRhaukus
  • UnkleRhaukus
\[ 5^{log{5}^{25}}\] ?
TuringTest
  • TuringTest
\[\huge 5^{(\log5)^{25}}\]?
UnkleRhaukus
  • UnkleRhaukus
is 5 the base of the logarithm ?
anonymous
  • anonymous
Turing Test wrote it correctly.
anonymous
  • anonymous
I know how to do it, just am confused about the 5 in the front. what do i do with it?
UnkleRhaukus
  • UnkleRhaukus
the the exponent of the log can come out the front \[5^{log(5)^{25}}=5^{25(log(5)}\]
anonymous
  • anonymous
Thats it? I am still confused.
TuringTest
  • TuringTest
\[5^{25\log5}\neq5^{25}\times5^{\log5}\]
UnkleRhaukus
  • UnkleRhaukus
I have made an error turing test?
TuringTest
  • TuringTest
\[\log_5(25)=2\]you are tired
anonymous
  • anonymous
I am reading this off of my math book. 5 is the big dog, while \[\log _{5}25\] is above it.
UnkleRhaukus
  • UnkleRhaukus
oh yeah i see how wrong that is now
TuringTest
  • TuringTest
well now that we've got it sorted out the problem is trivial goodman please try to post more clearly, that is not what you originally wrote
UnkleRhaukus
  • UnkleRhaukus
errors ervery where, perhaps i should just watch and learn
anonymous
  • anonymous
How did you get that? @TuringTest
TuringTest
  • TuringTest
\[\huge 5^{\log_5(25)}=5^2=25\]because\[\huge 5^2=25\]
anonymous
  • anonymous
I appologize, srry, i did read it wrong. srry srry.
anonymous
  • anonymous
I am new to log, so yea, srry.
TuringTest
  • TuringTest
but in general\[\huge a^{\log_a(x)}=x\]so we could have skipped that analaysis
anonymous
  • anonymous
Oh yea, because it always is equal to the front number. Yea, okay, got it :D Thanx, srry for the mix up :/
TuringTest
  • TuringTest
it's ok, let me know if you have more questions
anonymous
  • anonymous
You are familiar with log? Because dont understand em at all :P
TuringTest
  • TuringTest
yes, they can be a bit tricky at first think about small numbers first the definition as I like to give it:\[\huge\log_ax=b\text{ if }a^b=x\]in other words\[\large \log_ax\]asks "what power do we raise a to in order to get x?" for example...
TuringTest
  • TuringTest
\[\large\log_2(4)\]asks "2 raised to what power equals 4?" so what is the answer?
anonymous
  • anonymous
2 rite?
TuringTest
  • TuringTest
right because 2^2=4 so what about\[\log_2(16)\]?
anonymous
  • anonymous
4
TuringTest
  • TuringTest
right so now you should see why\[\log_5(25)=2\]yes?
anonymous
  • anonymous
YES!! wow!! It is 5 because 5^5 is equal to 25
TuringTest
  • TuringTest
\[5^2\]
anonymous
  • anonymous
Whoops..srry, so the exponent is the answer?
TuringTest
  • TuringTest
yeah\[\log_2(16)=4\]as you said, because\[2^4=16\]
anonymous
  • anonymous
Is that always the case?
TuringTest
  • TuringTest
so\[\log_5(25)=2\]because\[5^2=25\]
TuringTest
  • TuringTest
in general we have\[\huge\log_a(x)=y\iff a^y=x\]so yes, that means always you could say that the question is on the left: "a to what power equals x?" the answer is on the right "a to the y equals x"
TuringTest
  • TuringTest
so do some more what is\[\log_4(16)\]?
anonymous
  • anonymous
2
TuringTest
  • TuringTest
right\[\log_{10}(10000)\] ?
anonymous
  • anonymous
umm..lol, 1000
anonymous
  • anonymous
I think i got it wrong
TuringTest
  • TuringTest
the question is "10 to what power equals 10000" you are asserting that\[\large10^{1000}=10000\]which I dont think you believe what is\[10^2\] ?
anonymous
  • anonymous
Wait..no, its 100
TuringTest
  • TuringTest
\[10^{100}\neq10000\]
anonymous
  • anonymous
So what is it?
anonymous
  • anonymous
I dont have a calculator rite now :P
TuringTest
  • TuringTest
\[10^2=10\cdot10=100\]\[10^3=10\cdot10\cdot10=1000\]the exponent on the ten is the number of zeros after the one so 10 to what power equals 10000 ?
anonymous
  • anonymous
\[10^{4}?\]
TuringTest
  • TuringTest
exactly so\[\log_{10}(10000)=4\]because\[10^4=10000\]
anonymous
  • anonymous
That makes its it so simple
TuringTest
  • TuringTest
yeah that's a handy rule, and is used in science to describe very large and small numbers last one: how about\[\log_3(81)\]
anonymous
  • anonymous
4
TuringTest
  • TuringTest
right :D do you know the basic log rules\[\log(ab)=\log a+\log b\]\[\log(\frac ab)=\log a-\log b\]\[\log(a^b)=b\log a\]?
anonymous
  • anonymous
No, we havent been taught those. We started log in class today.
TuringTest
  • TuringTest
when you start using that things will be easier \[\large \log(ab)=\log a+\log b\text{ because }x^a\cdot x^b=x^{a+b}\]for example it takes a bit to get used to those ideas though
anonymous
  • anonymous
Yea, just by reading it, I understand the first function.
anonymous
  • anonymous
I also understand the second one, the third one looks funny.
TuringTest
  • TuringTest
the third one can be illustrated by the problems we just did first I will ask you what is\[\log_22\]?
anonymous
  • anonymous
1
anonymous
  • anonymous
lol, i find myself saying "2 raised to what power is 2" thats a neat trick
TuringTest
  • TuringTest
I'm glad, it helps me! and in general\[\log_aa=1\]now look at\[\log_2(16)\]if we factor 16 we get\[\log_2(2^4)\]now apply the third rule\[4\log_2(2)=4(1)=4\]so we get the answer we already knew it also shows that\[\log_aa^x=x\]which is nice to know that's how I did your earlier problem) so by factoring and using these rules we can break down lots of numbers with logs
TuringTest
  • TuringTest
now here's the last grand example\[\log_2(24)=\log_2(2^3\cdot3)\]using the first rule\[\log_2(2^3)+\log_23\]and now the third\[3\log_2(2)+\log_23=3+\log_23\]and that's as far as it goes
anonymous
  • anonymous
Wow, i never knew log could be so simple :D
TuringTest
  • TuringTest
I'm glad you feel that way :D
anonymous
  • anonymous
Thanx a ton :D I am now cleared up on log, thank you thank you!!!
TuringTest
  • TuringTest
happy to help!

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