SOLUTION: I'm having trouble with sovling the log equations. i have two questions. 3^(1-2x)=e^(0.5x). and log(x+1)+ log(x-1)=log(3) i know that you start off by taking the natural log of bot
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-> SOLUTION: I'm having trouble with sovling the log equations. i have two questions. 3^(1-2x)=e^(0.5x). and log(x+1)+ log(x-1)=log(3) i know that you start off by taking the natural log of bot
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Question 205449This question is from textbook Functions modeling Change
: I'm having trouble with sovling the log equations. i have two questions. 3^(1-2x)=e^(0.5x). and log(x+1)+ log(x-1)=log(3) i know that you start off by taking the natural log of both sides but after that i'm stuck. i don't know which logarithm property to you. your help is much appreciated This question is from textbook Functions modeling Change
You can put this solution on YOUR website! 3^(1-2x)=e^(0.5x)
ln(3)*(1-2x) = 0.5x
ln(3) - 2x*ln(3) = 0.5x
ln(3) = 0.5x + 2x*ln(3)
ln(3) = x*(0.5 + 2ln(3))
x = ln(3)/(0.5 + 2ln(3))
Everything on the right is a constant.
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log(x+1)+ log(x-1)=log(3)
log((x+1)*(x-1)) = log(3)
(x+1)*(x-1) = 3
x^2 - 1 = 3
x^2 = 4
x = +2
x = -2
The -2 won't work, as it gives log(-3), so
x = 2
