# Using the log properties- math

• 06-08-2009
ninety3gd
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...
• 06-08-2009
P4R4N01D
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...
• 06-08-2009
tabstop
Quote:

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...

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.
• 06-08-2009
ಠ_ಠ
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
• 06-08-2009
ಠ_ಠ
Quote:

Originally Posted by ninety3gd
how can I write b/a: Is it actually

logc (logcb - logxa) or

logc (X(1/Y)

logc(b/a) = logcb - logca, not logc(logcb - logca)
• 06-11-2009
ninety3gd
figured out my problem- thanks for all of the insight
• 06-11-2009
ಠ_ಠ
what was the solution?