Finding x when the exponents and coefficients are different How can I find $x$ when the exponents and coefficients are different?
$$3x \cdot e^{7x+2} = 15$$
Thanks
 A: When the variable you are solving for appears in both the exponent and as a coefficient, solving using the Lambert W Function is recommended.  To use the Lambert W Function, both coefficient and exponent must be the same and the base of the exponent must be $e$, Euler's number.
In other words, it must look like $ye^y=x$  It would then have the solution $y=W(x)$
$3xe^{7x+2}=15$
Divide by $e^2$ because it is difficult to solve most of these types of problems when extra numbers are being added in the exponent.
$3xe^{7x}=\frac{15}{e^2}=15e^{-2}$
Then multiply both sides of the equation by $\frac{7}{3}$ so that it becomes in the format of $ye^y=x$.
$7xe^{7x}=35e^{-2}$
You then take "W's" of both sides to solve.
$7x=W(35e^{-2})$
And then divide by 7.
$x=\frac{W(35e^{-2})}{7}$
As to a decimal value for the solution, there are Lambert W calculators that you can find online:  http://www.had2know.com/academics/lambert-w-function-calculator.html
Using this calculator...
$x\approx0.185148143$
A: In terms of the Lambert W Function, we have 
$$3xe^{7x+2}=15\implies 7xe^{7x}=35e^{-2}\implies x=\frac17 W\left(35e^{-2}\right)$$
