An Interesting Exponential Equation | Homemade

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[mcwulf]

Obvs x is negative so use plan A and divide by 6^x. Then flip each fraction and put y=-x:

(6/3)^y + (6/6)^y = (6/2)^y
2^y + 1 = 3^y
Clearly y=1 so x=-1.

[fadetoblah]

I think that dividing all terms by the larger of the three values (6^x) also works nicely. You end up with:
(1/2)^x + 1 = (1/3)^x, then
(1/3)^x - (1/2)^x = 1, and finally
3^-x - 2^-x = 1, where it becomes plain as day that x = -1 is a solution.
To demonstrate that it is the only solution, simply consider that:
- For positive values of x, 3^-x - 2^-x will always return negative values (so it can never equal 1)
- For negative values of x, 3^-x - 2^-x is a strictly decreasing function, such that it will intersect with y=1 exactly once, at x = -1.
Thanks as alway.

[billtruttschel]

Is there an algebraic way to solve this?

[TheRhythmOfMathematics]

This is typical way to solve this kind of exponential equation by using decreasing or increasing function.
To prove it as decreasing or increasing function, we can calculate the derivative and check the range of variable in the table.

[Skank_and_Gutterboy]

I divided all terms by 2^x to get (3/2)^x + 3^x = 1. Looking at the left side of the equation, I saw by inspection that it would equal 1 if both terms were "flipped". In other words, if you use x = -1, it works because: 2/3 + 1/3 = 1.

[someperson]

F(x) := (3/2)^x + 3^x, x real. F is the sum of two strictly increasing functions. Thus, F is strictly increasing.
Since F(-1) = 2/3 + 1/3 = 1, the only solution to F(x) = 1 is x = -1.
If 3^x + 6^x = 2^x, then, dividing by 2^x, yields, F(x) =1, proving that x = -1.

[jansentanu]

What's the app you use as the black board?

[mathswan]

f(x)=3^x+6^x-2^x
f(x) is increasing and f(-1)=0
Hence x=-1

[mbmillermo]

I think this is what people want to know: How do you solve a^x + b^x = c^x for x in terms of a, b and c? I doubt there is a way to get it without resorting to approximation for given a, b, c.