Please help me. find the real solution of the equation.

Question

asked Dec 28, 2022 56.5k views

1 Answer

SmallB · Answer 1 · 2023-01-02T02:44:52+0000

Answer:

$x = -5 \text{ or } x =3$

Explanation:

We want to find the real solutions to the equation:

$\displaystyle 8^(x^2 + 2x - 15) = 1$

Recall that any value raised to zero (except for zero itself) is one.

In other words, the exponent must equal zero:

x^2 + 2x - 15 = 0

Solve for x. Factor:

(x+5)(x-3) = 0

Zero Product Property:

$x + 5 = 0\text{ or } x - 3 = 0$

Solve for each case. Hence:

$x = -5 \text{ or } x =3$

In conclusion, our two real solutions are:

$x = -5 \text{ or } x =3$