SOLUTION: what is the anser to all are in bases of 5 log(base5) (5X-27) = log(base5) 12 - log(base5)3, could you show the work on solving it too? thank you!
Question 450326: what is the anser to all are in bases of 5 log(base5) (5X-27) = log(base5) 12 - log(base5)3, could you show the work on solving it too? thank you! Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! what is the answer to all are in bases of 5.
log(base5) (5X-27) = log(base5) 12 - log(base5)3
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log5(5x-27) = log5(12)-log5(3)
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log5(5x-27) = log5[12/3]
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log5(5x-27) = log5(4)
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5x-27 = 4
5x = 31
x = 31/5 = 6 1/5
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Cheers,
Stan H.
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