Using the log properties- math

Kind of an offshoot of my previous post but this has to do with logarithmic properties

Change of base:

logbX = logaX/logaB

and

logb(X/Y ) = logbX - logbY

and

X/Y = X(1/Y) exponents


here is the formula I have:
logc (b / a) where c = 3, b = 88 , a = 3

how can I write b/a: Is it actually

logc (logcb - logxa) or

logc (X(1/Y)

confused on how to proceed with the b/c- do I simpify that first then convert from base 5 to base 10

Insight would be appreciated...

How I learnt change of base: log of the number / log of the base. So to use a common base, to save me writing an extra 2 letters, and because my calculator can do it, I use natural log (ln). E.g.
with b = 88 , a = 3:
log3 (88) = ln (88) / ln (3) ~= 4.08

Or you could get a calculator to do it all for you...

Quote:

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Originally Posted by ninety3gd

Kind of an offshoot of my previous post but this has to do with logarithmic properties

Change of base:

logbX = logaX/logaB

and

logb(X/Y ) = logbX - logbY

and

X/Y = X(1/Y) exponents


here is the formula I have:
logc (b / a) where c = 3, b = 88 , a = 3

how can I write b/a: Is it actually

logc (logcb - logxa) or

logc (X(1/Y)

confused on how to proceed with the b/c- do I simpify that first then convert from base 5 to base 10

Insight would be appreciated...

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You don't have 5, or 10 in the problem. Do you want to do change of base, or the quotient rule? The quotient rule would say logc(b/a) = logc b - logc a. The change of base rule just allows you to find a numerical value with a calculator, should it be necessary in some way.

I wrote a program to convert a number in one base into another, it revolved around the modulus operator, I'll do some Googling and see if I can learn anything that can help you

Quote:

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Originally Posted by ninety3gd

how can I write b/a: Is it actually

logc (logcb - logxa) or

logc (X(1/Y)

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logc(b/a) = logcb - logca, not logc(logcb - logca)

figured out my problem- thanks for all of the insight

what was the solution?