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Help with logs?Cant use a calculator?:( Evaluate, without using a calculator 5^(-2)

Mathematics
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Oh wait i put the wrong one sowwy :( \[\log _{6}1\]
log(a)y = a^y
im a little confused.....would it be \[6^{1}\] than?

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Other answers:

yes :)
ohh okay thankyou(: um may you help me with another one? Its a fraction?
yesm
thankyou!(: okay here it is\[\log _{6}\frac{ 1 }{ \sqrt[5]{36}}\] sorry it took a while to type it lol
i did the first one wrong it should be log(a)x = y and a^y =x
ohh... okay im confused now :(
the second one is a little out of my league lol you should open a new question with that second part!
lol alright but may you explain the first one pleaseee?(: <3
so log(a)x is the exponent to which the base a must be raised to give x
so to raise 6 by a power of zero will give an output of 1. any number raised to zero will give 1
x^0=1
2^0=1 and so on
so.... okay im not 100% sure but would the answer be 1?
Oh and another question (sorry lol) so do you always raise the little number by 0?
log(6)1=0 6^0=1
nope you 'exponentiate' the little number
log(6)1=0 since 6^0=1
okay okay...what does exponentiate mean im sorry for all these questions
The act of raising a quantity to a power :)
okay(: but im still confused :'(
they confuse me too hehe
log(a)x = y <=> a^y = x
just remember that a is your base, the base will always be the same for any log
any number raised to a power like a^x, a will be your base
imm off gluck
alright:/ thanks though <3

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