SOLUTION: Can you check this answer please? Use the Properties of Logarithms to write as a single log: log12+(1/3)log7-log2 I know that adding means multiply them, so I have l

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Can you check this answer please? Use the Properties of Logarithms to write as a single log: log12+(1/3)log7-log2 I know that adding means multiply them, so I have l      Log On


   

Question 924208: Can you check this answer please?
Use the Properties of Logarithms to write as a single log:
log12+(1/3)log7-log2
I know that adding means multiply them,
so I have
log (12)*[7^(1/3)] all /2
Found 3 solutions by Alan3354, Theo, stanbon:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Use the Properties of Logarithms to write as a single log:
log12+(1/3)log7-log2
I know that adding means multiply them,
so I have
log (12) all /2
-----------
It's all /2, but inside the log
= log%2812%2Aroot%283%2C7%29%2F2%29
= log%286%2Aroot%283%2C7%29%29

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
i think you got it.

1/3 * log(7) = log(7^(1/3)

log(12) * log(7^(1/3) = log(12*7^(1/3) - log(2)

log(12*7^(1/3) - log(2) = log(12*7^(1/3)/2)

way to check is to calculate the answer from the original equation and the final equation.

the answer should be the same.

original equation gets 1.059850597

final equation gets the same.

final equation is good.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
log12+(1/3)log7-log2
-----
= log(12) + log(7^(1/3)) - log(2)
-----
= log[12*7^(1/3)/2]
-----
= log[6*7^(1/3)]
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Cheers,
Stan H.
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