SOLUTION: The given equations are quadratic. Solve analytically.
a) 3^2x + 35 = 12(3^x)
3^2x-36^x + 35 = 0
(3^x-35)(l^x-1)=0
not sure if I started this correctly or not.
Algebra ->
Exponential-and-logarithmic-functions
-> SOLUTION: The given equations are quadratic. Solve analytically.
a) 3^2x + 35 = 12(3^x)
3^2x-36^x + 35 = 0
(3^x-35)(l^x-1)=0
not sure if I started this correctly or not.
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Question 998824: The given equations are quadratic. Solve analytically.
a) 3^2x + 35 = 12(3^x)
3^2x-36^x + 35 = 0
(3^x-35)(l^x-1)=0
not sure if I started this correctly or not.
b) I was not sure how to do this one
2(lnx)^2 + 9 lnx =5
You can put this solution on YOUR website! a) 3^2x + 35 = 12(3^x)
(3^x)^2-36*3^x + 35 = 0
(3^x-35)(3^x-1)=0
3^x = 35 of 3^x = 1
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x = log(35)/log(3) or x = log(1)/log(3)
x = 1.8654 or x = 0
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b) I was not sure how to do this one
2(lnx)^2 + 9 lnx - 5 = 0
(2ln(x)-1)(2ln(x)+5) = 0
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ln(x) = 1/2 or ln(x) = -5/2
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x = e^(1/2) or x = e^(-5/2)
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Cheers,
Stan H.
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