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 ->
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
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(69443) (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.
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ln3x-ln15=2
Logs are exponents. Subtracting exponents --> dividing the arguments.
--> ln(3x/15) = 2
ln(x/5) = 2
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The base of ln's, natural logs, is e.
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email via the Thank You note if you have any questions.
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then i crossed out ln with e
divided the 15 & 3 =5x
and i got x=2 over 5 e
