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Write in exponential form. (Will write in equation editor)

Mathematics
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\[5^{\log5}^{25}\]
wait..that came out funny..
\[5^{\log5}\]

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Other answers:

There is suppoesd to be a 25 next to log 5
\[ 5^{log{5}^{25}}\] ?
\[\huge 5^{(\log5)^{25}}\]?
is 5 the base of the logarithm ?
Turing Test wrote it correctly.
I know how to do it, just am confused about the 5 in the front. what do i do with it?
the the exponent of the log can come out the front \[5^{log(5)^{25}}=5^{25(log(5)}\]
Thats it? I am still confused.
\[5^{25\log5}\neq5^{25}\times5^{\log5}\]
I have made an error turing test?
\[\log_5(25)=2\]you are tired
I am reading this off of my math book. 5 is the big dog, while \[\log _{5}25\] is above it.
oh yeah i see how wrong that is now
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
errors ervery where, perhaps i should just watch and learn
How did you get that? @TuringTest
\[\huge 5^{\log_5(25)}=5^2=25\]because\[\huge 5^2=25\]
I appologize, srry, i did read it wrong. srry srry.
I am new to log, so yea, srry.
but in general\[\huge a^{\log_a(x)}=x\]so we could have skipped that analaysis
Oh yea, because it always is equal to the front number. Yea, okay, got it :D Thanx, srry for the mix up :/
it's ok, let me know if you have more questions
You are familiar with log? Because dont understand em at all :P
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...
\[\large\log_2(4)\]asks "2 raised to what power equals 4?" so what is the answer?
2 rite?
right because 2^2=4 so what about\[\log_2(16)\]?
4
right so now you should see why\[\log_5(25)=2\]yes?
YES!! wow!! It is 5 because 5^5 is equal to 25
\[5^2\]
Whoops..srry, so the exponent is the answer?
yeah\[\log_2(16)=4\]as you said, because\[2^4=16\]
Is that always the case?
so\[\log_5(25)=2\]because\[5^2=25\]
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"
so do some more what is\[\log_4(16)\]?
2
right\[\log_{10}(10000)\] ?
umm..lol, 1000
I think i got it wrong
the question is "10 to what power equals 10000" you are asserting that\[\large10^{1000}=10000\]which I dont think you believe what is\[10^2\] ?
Wait..no, its 100
\[10^{100}\neq10000\]
So what is it?
I dont have a calculator rite now :P
\[10^2=10\cdot10=100\]\[10^3=10\cdot10\cdot10=1000\]the exponent on the ten is the number of zeros after the one so 10 to what power equals 10000 ?
\[10^{4}?\]
exactly so\[\log_{10}(10000)=4\]because\[10^4=10000\]
That makes its it so simple
yeah that's a handy rule, and is used in science to describe very large and small numbers last one: how about\[\log_3(81)\]
4
right :D do you know the basic log rules\[\log(ab)=\log a+\log b\]\[\log(\frac ab)=\log a-\log b\]\[\log(a^b)=b\log a\]?
No, we havent been taught those. We started log in class today.
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
Yea, just by reading it, I understand the first function.
I also understand the second one, the third one looks funny.
the third one can be illustrated by the problems we just did first I will ask you what is\[\log_22\]?
1
lol, i find myself saying "2 raised to what power is 2" thats a neat trick
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
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
Wow, i never knew log could be so simple :D
I'm glad you feel that way :D
Thanx a ton :D I am now cleared up on log, thank you thank you!!!
happy to help!

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