(x-3)^x+2-(x-3)^x+8=0

3 min read Jun 17, 2024

Solving the Equation (x-3)^x+2 - (x-3)^x+8 = 0

This equation might look daunting at first glance, but we can solve it using a bit of algebraic manipulation and some clever observations.

First, let's try to simplify the equation. Notice that both terms on the left-hand side share a common factor of (x-3)^x. We can factor this out:

(x-3)^x * [(x-3)^2 - (x-3)^8] = 0

Now we have a product of two terms equal to zero. This means at least one of the terms must be equal to zero:

The equation (x-3)^x = 0 is only satisfied when x = 3. This is because any non-zero number raised to a power will never be equal to zero.

The second term requires a bit more work. Let's make a substitution:

Now our equation becomes:

y^2 - y^8 = 0

We can factor out a y^2:

y^2 * (1 - y^6) = 0

This gives us two possibilities:

y^2 = 0 This means y = 0, which in turn means x = 3.
1 - y^6 = 0 This means y^6 = 1. The solutions to this equation are y = 1 and y = -1. This translates to x = 4 and x = 2.

Therefore, the solutions to the original equation (x-3)^x+2 - (x-3)^x+8 = 0 are: