Multiply by a^x throughout:
Therefore, completing the square;
Rearrange for a^x;
Ok, that's as far as I'm sure for this part. For the other one, I'm not quite getting something.
If z = a^3x = a^-3x;
a^x = z^1/3 = z^-1/3.
Taking a^x = z^1/3;
Expanding, I get:
Maybe it's because I didn't take a^x = z^-1/3, but if I do so, I'll get infinite terms, and I don't know how to deal with that.
Is that it? If so, multiply throughout by (x+1)^1/2;
You can make use of a substitution to help you solve here. Let (x+1)^1/2 = y;
So, 3y -5 = 0,
y = 5/3
and y +1 = 0
y = -1
You can solve those two now.
For the first part, I prefer not answering because I don't quite understand what the question is asking for. I can solve for x, and get x = 2, but not show that it's a real root, sorry