Question 842765: How do you condense 4log, base 9, of 7-4log, base 9, of 7- 12log, base of 9, 12 to a single logarithm?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! I believe you are talking about:
4 * log9(7) - 4 * log9(7) - 12 * log9(12)
you will use a couple of peoperties of logarithms.
those are:
log(a^b) = a*log(b)
log(a/b) = log(a) - log(b)
apply these to your problem as follows:
4 * log9(7) - 4 * log9(7) - 12 * log9(12)
becomes:
log9(7^4) - log9(7^4) - log9(12^12)
because a * log(b) = log(b^a)
this becomes:
log9(7^4/7^4/12^12)
because log(a) - log(b) - log(c) = log(a/b/c)
since these are all numbers, you can check to see if you did it correctly by applying the rules of logs to both equations.
log9(7^4 / 7^4 / 12^12) is equal to LOG(7^4 / 7^4 / 12^12) / LOG(9) which is equal to -13.57115704
4log9(7) - 4log9(7) - 12log9(12) is equal to:
4LOG(7)/LOG(9) - 4LOG(7)/LOG(9) - 12LOG(12)/LOG(9) which is equal to -13.57115704
Both the initial log equation and the final log equation give the same answer so the conversion is good.
the log base conversion formula is:
logb(x) = logc(x) / logc(b)
when using the calculator, are either converting to the base 10 (LOG function) or to the base of e (LN function).
either one will get you a correct answer.
I used conversion to LOG function.
I could just as easily have used conversion to LN function.
your converted formula that you are looking for is:
log9(7^4/7^4/12^12)
|
|
|
