Question

Solve $(x²+1)^{(x+5)} = (3x+1)^{(x+5)}$

Answer 1

x = -5, 0, 3

Answer 2

To solve the equation $(x^2+1)^{x+5} = (3x+1)^{x+5}$, we consider two cases:

Case 1: When the exponent is zero

If $x+5 = 0$, then $x = -5$. In this case, both sides of the equation become 1, since any non-zero number raised to the power of 0 is 1. So, $(x^2+1)^0 = (3x+1)^0 = 1$, provided the bases are nonzero. Since $x^2+1 > 0$ for all $x$ and $3(-5)+1 = -14 \neq 0$, $x=-5$ is a valid solution.

Case 2: When the exponent is non-zero

If $x+5 \neq 0$, then we can equate the bases:

$$x^2+1 = 3x+1$$

Subtracting 1 from both sides:

$$x^2 = 3x$$

Rearranging the equation:

$$x^2 - 3x = 0$$

Factoring out x:

$$x(x-3) = 0$$

So, the solutions are $x = 0$ or $x = 3$.

Therefore, the solutions to the equation are $x = -5, 0, 3$.