Solving An Interesting Exponential Equation

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[Don-Ensley]

problem
2ˣ^³ = 3ˣ^²

Take natural logarithms on each side and bring down exponents.

x³ ln 2 = x² ln 3

Subtract.

x³ ln 2 - x² ln 3 = 0

Factor.

x² ( x ln 2 - ln 3 ) = 0

By zero product property,
x = 0
This is when both sides of the original equation are 1. 2ˣ^³ is interesting since the exponent is positive when x > 0 where the function increases exponentially and the exponential is negative when x < 0 where the function decreases towards 0 asymptomatically. 3ˣ^² on the other hand has a positive exponentially if x is positive or negative so it takes off exponentially in either direction giving it a u shaped curve that bottoms out at x = 0.

The other factor gives us
x = ln 3 / ln 2

, which is the second intersection.

Define f(x) = 2ˣ^³ and g(x) = 3ˣ^².

Then
f'(x) = 2ˣ^³ ln 2 (3 x²)
f"(x) = 2ˣ^³ 3x ln 2 [ 3 x³ ln 2 + 2 ]
f(x) has a local maximum at x= 0 which is also an inflection point.
It is interesting that f(x) goes through 2 inflection points. It starts out on the interval (-∞, 0] opening upwards then at

x = - ∛ ( 2 /(ln 8) )

It starts opening downwards. Then at x= 0 to ∞ it opens upwards again. It is cool to see how the exponential reflects the cubic in the exponent!

g'(x) = 3ˣ^² ln 3 (2x)
g"(x) = 3ˣ^² 2 ln 3 ( 2 x² ln 3 + 1 )

This function opens upwards everywhere, with the slope negative below 0 and positive above 0. The minimum is at x = 0 at g(0) = 1, the intersection with f(0).

f(x) drops away from g(x) in the negative direction but increases towards g(x) in the positive x direction. At first g(x) increases faster than f(x) then at x = ln3/ln2 f(x) is at a faster rate than g(x) with

f'(ln 3/ ln 2) = 2^((ln 3/ ln 2)³ ) ln 2 (3 (ln 3/ ln 2)²)
( approx 82.520 )

g'(ln 3/ ln 2) = 3^((ln 3/ ln 2)² ) ln 3 (2 ln 3/ ln 2)
( approx 55.014 )

answer

x ∈ { 0, ln 3 / ln 2 }

[renesperb]

The solution x= 0 is obvious.If you take logarithms you get x^3 ln 2 = x^2 ln3, that is x= ln3/ln2.

[gelbkehlchen]

Solution:
2^(x³) = 3^(x²) |ln() ⟹
x³*ln(2) = x²*ln(3) |-x²*ln(3) ⟹
x³*ln(2)-x²*ln(3) = x²*[x*ln(2)-ln(3)] = 0
|either x² = 0 ⟹ x = 0 or
x*ln(2)-ln(3) = 0 |+ln(3) ⟹
x*ln(2) = ln(3) |/ln(2) ⟹
x = ln(3)/ln(2) ≈ 1, 5850