Does anyone know how you go about solving the following equation for x:
ln(x)+k1*x=k2
where k1 and k2 are constants?
ln(x)+k1*x=k2
where k1 and k2 are constants?
x = K2/K1 =1 is an algebraic solutionGet all the logarithms on one side of the equation:
\(
\ln (x)= k_2-k_1 x
\)
and take the exponent
\(
x = e^{k_2-k_1 x}
\)
\(
x=\frac{e^{k_2}}{e^{k_1 x}}
\)
\(e^{k_2}\) is a constant, so call it K
\(
x=\frac{K}{ e^{k_1 x}}
\)
and I don't think it has an algebraic solution.
I know - I hit "Post Reply" instead of "Preview". I've edited it now, and I'll put "[Edit]" when I've checked it.You dropped a minus sign
I realized that as I was trying to show the correctionI know - I hit "Post Reply" instead of "Preview". I've edited it now, and I'll put "[Edit]" when I've checked it.
What are the values of k1 and k2? If they are unknown constants, then they are variables. With three variables, you can get to any value.Does anyone know how you go about solving the following equation for x:
ln(x)+k1*x=k2
where k1 and k2 are constants?
x = K2/K1 =1 is an algebraic solution
They are unknown 'constants' where we are trying to solve for x. That is true.What are the values of k1 and k2? If they are unknown constants, then they are variables.
k1=k2=k:x = K2/K1 =1 is an algebraic solution
by Jake Hertz
by Ikimi .O
by Ikimi .O