Equation problem

Here's another way to approach this question:

The equation can be written as 3x = e^-x

e^-x can be written as a series: 1 - x + x²/2 - x³/6 + x⁴/24 - x⁵/120 + ...

So 3x = e^-x can be written as 3x = 1 - x + x²/2 - x³/6 + x⁴/24 - x⁵/120 + ...

Subtract 3x from both sides to get  0 = 1 - 4x + x²/2 - x³/6 + x⁴/24 - x⁵/120 + ...

Now approximate this by  0 ≈ 1 - 4x₀  so that x₀ ≈ 0.25  

Take the next term to get   0 ≈ 1 - 4x₁ + x₁²/2   Solve this to get x₁ ≈ 0.258  (ignoring the other solution which is bigger than 1, so the series would diverge, and is clearly not a solution to the original equation).

The true solution lies between 0.25 and 0.258  (the question allows for an interval as a solution!).

It is easy enough to take higher order approximations numerically to tighhten the interval if required.

.

x=(3*e^x)^-1

1/x=3*e^x

1/3x=e^x

ln(1/3x) = x

-ln(3)-ln(x)=x
-ln(3) = x + ln(x)

-(log(3)/log(e))=-1.0986122886681098 = x + ln(x)

This is a hard problem, I hope I may have given you enough to solve it, but I forget how to do this kind of thing.

x=1/(3*e^x)

The real value can be obtained either through graphing or interpolation. My calculator gives a value of x=0.2576276530.......through interpolation.

Here's a graphical solution : https://www.desmos.com/calculator/c0vu5ffqbv

It  occurs  at about  x = 0.189

CPhill: The figure of .189 does not appear to balance the equation? The figure that I obtained, x=0.2576276530, does to a dozen decimal places.

Sorry......I typed the wrong function into Desmos......Guest 2  is correct.....the answer occurs at about x = 0.258

Here's the correct graph ........https://www.desmos.com/calculator/t9ttiydcve

x=1/(3*e^x)

solve for x (or give an interval that satisfies the equation)

We use the product log function, also called Lambert W-Function:

If the function is $x\cdot e^x = z$, then $x = W(z)$

$\begin{array}{rcll} x &=& \dfrac{1} {3\cdot e^x} \qquad & | \qquad \cdot e^x \\ x\cdot e^x &=& \frac13\\ \mathbf{ x } & \mathbf{=} & \mathbf{ W(\frac13) } \end{array}$

$x = 0.257627653049736704282916201626097790909692647503204491533\dots$

Nice, Alan  !!!!