Question 978650: Solve: 15(3^x + 1) - 243(5^x - 2) = 0
(Exponential and logarithmic functions)
Please, please help me with this question! Thank you!! Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! 15(3^x + 1) - 243(5^x - 2) = 0
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What are the exponents?
3^x or 3^(x+1) ?
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The exponents for my question (Question 978650) are 3^(x+1) and 5^(x-2).
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Then it should read:
15*3^(x + 1) - 243*5^(x - 2) = 0
log(15) + (x+1)*log(3) - log(243) - (x-2)*log(5) = 0
x*log(3) - x*log(5) + log(15) + log(3) - log(243) + 2log(5) = 0
x*(log3/5) + log(15*3*25/243) = 0
x*log(0.6) + log(125/27) = 0
x = -log(0.6)/log(125/27)
x =~ 0.333333
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x = 1/3
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There's probably a better method, I'll work on it later.
15*3^(x + 1) - 243*5^(x - 2) = 0
45*3^x - 243*5^x/25 = 0
5*3^x - 27*5^x/25 = 0
125*3^x - 27*5^x = 0
log(125) + x*log(3) - log(27) - x*log(5) = 0
x*log(3) - 3log(3) = x*log(5) - 3log(5)
