Answer:
[tex]x = -5 \text{ or } x =3[/tex]
Step-by-step explanation:
We want to find the real solutions to the equation:
[tex]\displaystyle 8^{x^2 + 2x - 15} = 1[/tex]
Recall that any value raised to zero (except for zero itself) is one.
In other words, the exponent must equal zero:
[tex]x^2 + 2x - 15 = 0[/tex]
Solve for x. Factor:
[tex](x+5)(x-3) = 0[/tex]
Zero Product Property:
[tex]x + 5 = 0\text{ or } x - 3 = 0[/tex]
Solve for each case. Hence:
[tex]x = -5 \text{ or } x =3[/tex]
In conclusion, our two real solutions are:
[tex]x = -5 \text{ or } x =3[/tex]
