Question 1189364: 1.Solve the following equations:
(i) ln x + ln(x + 4) = ln(x + 6)
(ii) 7^x+3 = e^x
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 1.Solve the following equations:
(i) ln x + ln(x + 4) = ln(x + 6)
ln(x*(x+4)) = ln(x+6)
x^2 + 4x = x+6
x^2 + 3x = 6
x^2 + 3x + 2.25 = 8.25
(x + 1.5)^2 = 8.25
x+1.5 = sqrt(33)/2, -sqrt(33)/2
x = (-3 + sqrt(33))/2 =~ 1.37228
The other "solution" is negative and is not valid due to the logarithm.
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(ii) 7^x+3 = e^x
Is x the exponent of 7?
Or x+3?
If it's x+3:
----------------
7^(x+3) = e^x
ln(7^(x+3)) = ln(e^x)
(x+3)*ln(7) = x
x*ln(7) + 3ln(7) = x
x*ln(7) + ln(343) = x
x(ln(7) - 1) = -ln(343)
x = -ln(343)/(ln(7) - 1)
x =~ -6.1715
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Check:
e^x =~ 2.0881e-3 ---- (e-3 is scientific format *10^-3)
7^-3.1715 =~ 2.0882e-3
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