# Thread: solving this with ln

1. ## solving this with ln

A bit rusty on the ln's.

1.5=ke^3k.

attempt

ln1.5=3klne

ln1.5=3k(1)

(ln1.5)/3=k???

Thanks....

2. ## Re: solving this with ln

Originally Posted by Neverquit
A bit rusty on the ln's.

1.5=ke^3k.

attempt

ln1.5=3klne

ln1.5=3k(1)

(ln1.5)/3=k???

Thanks....
you have k as a coefficient of the exponential term and k as part of the exponent. An elementary algebraic approach to solve for k is not available ... you'll need to use technology.

let $\displaystyle k = x$

graph $\displaystyle y = xe^{3x} - 1.5$ and locate any zeros ...

$\displaystyle x = k = 0.4224126...$

3. ## Re: solving this with ln

Originally Posted by Neverquit
A bit rusty on the ln's.

1.5=ke^3k.

attempt

ln1.5=3klne

ln1.5=3k(1)

(ln1.5)/3=k???

Thanks....
I will assume that equation is :

$\displaystyle 1.5=k\cdot e^{3k}$

$\displaystyle 4.5 = 3k \cdot e^{3k} \Rightarrow 3k=W(4.5) \Rightarrow k=\frac{W(4.5)}{3}$

where $\displaystyle ~W~$ is Lambert W function .