# SOLUTION: Hi, I've been having trouble with logarithms, i tried to solve the problem below but i'm not sure if it's correct. ln3x-ln15=2 How i tried to solve: ln3x-ln15=2 ln3x div

Algebra ->  Algebra  -> Exponential-and-logarithmic-functions -> SOLUTION: Hi, I've been having trouble with logarithms, i tried to solve the problem below but i'm not sure if it's correct. ln3x-ln15=2 How i tried to solve: ln3x-ln15=2 ln3x div      Log On

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 Question 590108: Hi, I've been having trouble with logarithms, i tried to solve the problem below but i'm not sure if it's correct. ln3x-ln15=2 How i tried to solve: ln3x-ln15=2 ln3x divided by ln 15 = 2 then i crossed out ln with e divided the 15 & 3 =5x and i got x=2 over 5 e Hopefully, that made sense.Answer by Alan3354(30993)   (Show Source): You can put this solution on YOUR website!ln3x-ln15=2 How i tried to solve: ln3x-ln15=2 ln3x divided by ln 15 = 2 It says minus, you can't change it to divide. ----------------------------- ln3x-ln15=2 Logs are exponents. Subtracting exponents --> dividing the arguments. --> ln(3x/15) = 2 ln(x/5) = 2 --------------- The base of ln's, natural logs, is e. ---------------- ================ email via the Thank You note if you have any questions. -------------------------------------------------------- then i crossed out ln with e divided the 15 & 3 =5x and i got x=2 over 5 e