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How do you solve this equation? Does it contain log?
10^(18-x) = 26^x
x^2e^x + 5xe^x = 14e^x
e^2x-5e^x=36
These are three examples of the 50 problems that I have to do. If I know how to do these three, I can basically do everything else myself.
2 Answers
- 1 decade agoFavorite Answer
(18 - x) * ln(10) = x * ln(26)
18 * ln(10) = x * ln(26) + x * ln(10)
18 * ln(10) = x * (ln(260))
ln(10^18) / ln(260) = x
x^2 * e^x + 5x * e^x = 14 * e^x
e^x * (x^2 + 5x - 14) = 0
e^x * (x + 7) * (x - 2) = 0
e^x = 0
x = ln(0)
ln(0) does not exist
x = 2 , -7
e^(2x) - 5 * e^x - 36 = 0
e^x = (5 +/- sqrt(25 + 4 * 36)) / 2
e^x = (5 +/- sqrt(169)) / 2
e^x = (5 +/- 13) / 2
e^x = 18/2 , -8/2
e^x = 9 , -4
e^x cannot equal -4
e^x = 9
ln(9) = x
- Ahmed AbaLv 61 decade ago
1) take log to both
(18-x)log10=xlog26
18log10-xlog10=xlog26
18log10 =xlog26+xlog10
18log10=x(log26+log10)
x=(18log10)/(log10+log26)=18log10/log260
2)x^2.e^x+5x.e^x-14e^x=0
e^x(2x^2+5x-14)=0
e^x(x+7)(x-5)=0 then x=-7,x=5 since e^x cant be 0
3)take LN to both you get
2x(Ln e) -x(Ln 5e)=Ln 36
2x-xLn5 -xLn e=Ln36 while Ln e=1 (Ln5e=Ln5+Lne)
2x-xLn5-x=ln36
x-xLn 5=Ln36
x(1-Ln5)=Ln36
x=Ln36/(1-Ln5)
