SOLUTION: I was told "Write in condensed form (as a single logarithm)"
log2x+3log3(x-1)-log3(3-2x)
Log base 2 of x plus 3 log base 3 of x-1 minus log base 3 of 3-2x.
So far I have d
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-> SOLUTION: I was told "Write in condensed form (as a single logarithm)"
log2x+3log3(x-1)-log3(3-2x)
Log base 2 of x plus 3 log base 3 of x-1 minus log base 3 of 3-2x.
So far I have d
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Question 441593: I was told "Write in condensed form (as a single logarithm)"
log2x+3log3(x-1)-log3(3-2x)
Log base 2 of x plus 3 log base 3 of x-1 minus log base 3 of 3-2x.
So far I have done...
(Log x)/(log 2) + (log (x-1) ^3)/(log 3) – (log (3 – 2x))/(log 3)
(Log of x over log of 2) plus (log of x-1 over log of 3) minus ( log of 3 - 2x over log of 3)
Then...
Log ((x(x-1))/6) - ((3 - 2x)/3)
And finally I came up with
log [(x^2 - x)/6]/[(3 - 2x)/3]
I KNOW this is wrong. I am having trouble because two of the logs have the same base and one of the logs has a different base. I'm assuming that logs work like common terms (like 2x^2 can be added with x^2) and cannot be combined because they are different. I was given this problem on a take home test and a problem with two different logs has not appeared in my book or my notes at all. I just really need help with this equation! Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! log2x+3log3(x-1)-log3(3-2x)
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= log2(x) + log3(x-1)^3 - log3(3-2x)
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= log2(x) + log3[(x-1)^3/(3-2x)]
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Working with the base 2 term:
log2(x) can be written as log3(x)/log3(2) = (1/log3(2))log3(x)
= (log(3)/log(2))log3(x) = 1.5850*log3(x) = log3(x^1.5850)
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Getting back to the problem:
= log3(x^1.5850)+log3[(x-1)^3/(3-2x)]
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Final: log3[(x^1.5850*(x-1)^3)/(3-2x)]
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Cheers,
Stan H.
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