Precalculus logarithm question

  • Thread starter gusty987
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I'm trying to use LOGs other than log base 10 and base e on my TI-86. Can I accomplish this like this?:

log base a of b = (log base 10 of b) / (log base 10 of a) ?

or is it:

log base a of b = (ln b) / (ln a) ?

Help needed ASAP. Thanks!
 
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Both are correct.
 

Galileo

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I'll use [itex]^a\log b[/itex] for the log base a of b.

For any positive a,b:
[tex]^a\log b=x \iff a^x=b \iff \ln a^x=\ln b \iff[/tex]
[tex] x\ln a = \ln b \iff x=\frac{\ln b}{\ln a}[/tex]

I took the natural logarithm, but that was a complety arbitrary.
Therefore:

[tex]^a\log b=\frac{\ln b}{\ln a}=\frac{^{10}\log b}{^{10}\log a}=\frac{^y\log b}{^y\log a}[/tex]
for any base y.
 
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