Math - natural log
posted by Anonymous .
Solve for x. (25e^0.1x)/(e^(0.1x) + 3) = 10
e^2.5x / (e^(0.1x) + 3) = 10
e^(2.5x - 0.1x) + 1/3 = 10
e^2.4x = 29/3
ln^(e^(2.4x)) = ln(29/3)
2.4x = ln(29/3)
x = 0.945
Is that correct?
posted by Anonymous .
Solve for x. (25e^0.1x)/(e^(0.1x) + 3) = 10
e^2.5x / (e^(0.1x) + 3) = 10
e^(2.5x - 0.1x) + 1/3 = 10
e^2.4x = 29/3
ln^(e^(2.4x)) = ln(29/3)
2.4x = ln(29/3)
x = 0.945
Is that correct?