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

- anonymous

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

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- schrodinger

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- anonymous

\[5^{\log5}^{25}\]

- anonymous

wait..that came out funny..

- anonymous

\[5^{\log5}\]

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## More answers

- anonymous

There is suppoesd to be a 25 next to log 5

- UnkleRhaukus

\[ 5^{log{5}^{25}}\] ?

- TuringTest

\[\huge 5^{(\log5)^{25}}\]?

- UnkleRhaukus

is 5 the base of the logarithm ?

- anonymous

Turing Test wrote it correctly.

- anonymous

I know how to do it, just am confused about the 5 in the front. what do i do with it?

- UnkleRhaukus

the the exponent of the log can come out the front
\[5^{log(5)^{25}}=5^{25(log(5)}\]

- anonymous

Thats it? I am still confused.

- TuringTest

\[5^{25\log5}\neq5^{25}\times5^{\log5}\]

- UnkleRhaukus

I have made an error turing test?

- TuringTest

\[\log_5(25)=2\]you are tired

- anonymous

I am reading this off of my math book. 5 is the big dog, while \[\log _{5}25\] is above it.

- UnkleRhaukus

oh yeah i see how wrong that is now

- 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

errors ervery where,
perhaps i should just watch and learn

- anonymous

How did you get that? @TuringTest

- TuringTest

\[\huge 5^{\log_5(25)}=5^2=25\]because\[\huge 5^2=25\]

- anonymous

I appologize, srry, i did read it wrong. srry srry.

- anonymous

I am new to log, so yea, srry.

- TuringTest

but in general\[\huge a^{\log_a(x)}=x\]so we could have skipped that analaysis

- anonymous

Oh yea, because it always is equal to the front number. Yea, okay, got it :D Thanx, srry for the mix up :/

- TuringTest

it's ok, let me know if you have more questions

- anonymous

You are familiar with log? Because dont understand em at all :P

- 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

\[\large\log_2(4)\]asks
"2 raised to what power equals 4?"
so what is the answer?

- anonymous

2 rite?

- TuringTest

right
because 2^2=4
so what about\[\log_2(16)\]?

- anonymous

4

- TuringTest

right
so now you should see why\[\log_5(25)=2\]yes?

- anonymous

YES!! wow!! It is 5 because 5^5 is equal to 25

- TuringTest

\[5^2\]

- anonymous

Whoops..srry, so the exponent is the answer?

- TuringTest

yeah\[\log_2(16)=4\]as you said, because\[2^4=16\]

- anonymous

Is that always the case?

- TuringTest

so\[\log_5(25)=2\]because\[5^2=25\]

- 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

so do some more
what is\[\log_4(16)\]?

- anonymous

2

- TuringTest

right\[\log_{10}(10000)\] ?

- anonymous

umm..lol, 1000

- anonymous

I think i got it wrong

- 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

Wait..no, its 100

- TuringTest

\[10^{100}\neq10000\]

- anonymous

So what is it?

- anonymous

I dont have a calculator rite now :P

- 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

\[10^{4}?\]

- TuringTest

exactly
so\[\log_{10}(10000)=4\]because\[10^4=10000\]

- anonymous

That makes its it so simple

- 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

4

- 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

No, we havent been taught those. We started log in class today.

- 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

Yea, just by reading it, I understand the first function.

- anonymous

I also understand the second one, the third one looks funny.

- TuringTest

the third one can be illustrated by the problems we just did
first I will ask you what is\[\log_22\]?

- anonymous

1

- anonymous

lol, i find myself saying "2 raised to what power is 2" thats a neat trick

- 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

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

Wow, i never knew log could be so simple :D

- TuringTest

I'm glad you feel that way :D

- anonymous

Thanx a ton :D I am now cleared up on log, thank you thank you!!!

- TuringTest

happy to help!

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