Question 614936: What is 3log x+6 log (x-2) written as a single logarithm?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 3*log(x) + 6*log(x-2) would be equivalent to:
log(x^3) + log((x-2)^6) which would then be equivalent to:
log(x^3 * (x-2)^6)
the 2 properties of logarithms that are used are:
log(a*b) = log(a) + log(b)
log(a^b) = b*log(a)
you need to apply them in the correct order.
assume you started with:
log(x^3 * (x-2)^6)
you would use log(a*b) = log(a) + log(b) to get:
log(x^3) + log(x-2)^6)
you would then use log(a^b) = b*log(a) to get:
3*log(x) + 6*log(x-2)
to confirm you have the correct answer, substitute random values for x and solve using the original equation and the final equation to see if you get the same answer with each.
assuming x is equal to 9, solve each in turn.
original equation:
3*log(x) + 6*log(x-2) is equal to:
3*log(9) + 6*log(9-2) which is equal to:
3*log(9) + 6*log(7) which is equal to:
7.933315768
final equation:
log(x^3 * (x-2)^6) is equal to:
log(9^3 * (9-2)^6) which is equal to:
log(9^3 * 7^6) which is equal to:
log(729 * 117649) which is equal to:
log(85766121) which is equal to:
7.933315768
the answer are the same so you can reasonably assume that the conversion was done successfully.
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