# SOLUTION: Thank you so much nerdybill for helping me with my previous problem, Question 169324. I was raising 1, to the 9th power, which is 1 - instead of 9 to the 1st power. I have a sec

Algebra ->  Algebra  -> Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Thank you so much nerdybill for helping me with my previous problem, Question 169324. I was raising 1, to the 9th power, which is 1 - instead of 9 to the 1st power. I have a sec      Log On

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 Question 169385: Thank you so much nerdybill for helping me with my previous problem, Question 169324. I was raising 1, to the 9th power, which is 1 - instead of 9 to the 1st power. I have a second question, I can't seem to wrap my mind around. How do you solve: log base a 5 = 0.699; log base a 3 = 0.477; find log base a 15 rewritten: log (a,5)=0.699; log (a, 3) = 0.477; find log (a,15).Answer by Alan3354(30993)   (Show Source): You can put this solution on YOUR website! How do you solve: log base a 5 = 0.699; log base a 3 = 0.477; find log base a 15 rewritten: log (a,5)=0.699; log (a, 3) = 0.477; find log (a,15). ------------------- The log of a product is the sum of the logs of its elements. log(15) = log(3) + log(5), IF they all have the same base. So, the log(15) is 0.699 + 0.477 log(15) = 1.176