Precalculus logarithm question

[gusty987]

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!

[Muzza]

Both are correct.

[Galileo]

I'll use ^a\log b for the log base a of b.

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

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

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