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Logarithms

hey how do you find the value of:

loga(1)???
Original post by kickboxer 98
hey how do you find the value of:

loga(1)???


Use the definition of the logarithm, find xx such that ax=1a^x = 1.
Original post by EricPiphany
Use the definition of the logarithm, find xx such that ax=1a^x = 1.


i thought the answer was 1 but i dont know?
Anything to the power of 0 = 1
Original post by kickboxer 98
i thought the answer was 1 but i dont know?


The answer isn't 1. What do you know about the log function?
Reply 5
Avatar for Andy98
Andy98
20
Original post by kickboxer 98
hey how do you find the value of:

loga(1)???


ax=1a^x=1

What is x?

EDIT: They beat me to it :frown:
(edited 8 years ago)
is the answer 0?
anything to the power of 0 is 1?
Reply 7
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Shad123
Original post by Bobjim12
Anything to the power of 0 = 1


thats what i thought
Reply 8
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Zacken
22
Original post by kickboxer 98
is the answer 0?
anything to the power of 0 is 1?


Yes, precisely! :-)
okay, now... if i had:

2(loga 6) -1

how would that expand to remove the brakets?
Reply 10
Avatar for Zacken
Zacken
22
Original post by kickboxer 98
okay, now... if i had:

2(loga 6) -1

how would that expand to remove the brakets?


2log⁡a6=log⁡a622\log_a 6 = \log_a 6^2
Original post by Zacken
2log⁡a6=log⁡a622\log_a 6 = \log_a 6^2


what happened to the -1?

question is:
loga X = 2(loa 6)-1

find X
Reply 12
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Zacken
22
Original post by kickboxer 98
what happened to the -1?

question is:
loga X = 2(loa 6)-1

find X


I was only simplifying the log term for you.

You want to get both sides in terms of log⁡a\log_a so you can 'cancel' it out.

log⁡ax=log⁡a62−log⁡aa=log⁡a62a\log_a x = \log_a 6^2 - \log_a a = \log_a \frac{6^2}{a}
Reply 13
Avatar for AG98
AG98
1
I suggest you familiarise yourself with the log rules a bit before you attempt the questions you are currently working on :smile:

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