Precalculus 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!

 

[DWW]

You may find better help over at the physicsforums.com forums. There are many different forums (not just physics) including math help.

Lots of smart people there 😉

 

[DrPizza]

I never have students memorize the formula you're trying to remember for that exact reason...
it's hard to remember which way it goes.

So, if you have log base a of b, you can write it as
x = log base a of b
rewrite it in exponent form
a^x=b
Then, log both sides
log a^x = log b

Use the exponent property of logs to move the x to the coefficient spot
x log a = log b

And finally, solve for x again...

x = log b / log a

And, there you have it.
(log without base written is intended to be log base 10)

 

[DrPizza]

From my derivation of the formula, you can see that you can also use ln b / ln a
Or a log of any other base - if you had buttons on the calculator for it 😛

 

[cw42]

wow... i've fried my brain all summer and i don't remember anything i learned my first year of college. another month to waste here we go august!

 

[Gusty987]

Thanks, DrPizza!