Can someone help me with exponential equations?

Question

ok.. I think if someone can show me step for step how to do this I can figure out the rest of the problems

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

please help!!!

thank you!!! man...that was making my brain hurt.... lol.

Answer 1

rewrite the equation
(3^x)^2 + 3^x - 2 = 0

let y be 3^x. Substitute
y^2 + y - 2 = 0

factor
(y + 2) (y - 1) = 0

y = 1 or -2

we know that y = 3^x

so 1 = 3^x
x = log(1)/log(3)
x = 0

-2 = 3^x
x = log(-2)/log(3)
x = no root

so the only answer is 0

Answer 2

Substitute u for 3^x and the equation becomes:

u^2 + u - 2 = 0

You can then solve by factoring:
(u-1)(u+2) = 0
so u = 1 or u = -2

Put 3^x back in:

3^x = 1 or 3^x = -2

The 2nd equation is impossible, so 3^x = 1 leads to x = 0 since any # to the zero power = 1.

Hope that helps!

Answer 3

3^(2x) + 3^(x) - 2 = 0
(3^x)^2 + 3^x - 2 = 0 (Factorise 3^x is a common factor)
Let 3^x = y

Then
y^2 + y - 2 = 0
(y + 2)(y - 1) = 0
Hence y = -2 and y = +1
It follows 3^x = -2 and 3^x = 1

Take logs.
log(10)3^x = log(10)1
xlog(10)3 = log(10)1
x(0.4771) = 0
x = 0

log(10)3^x = log(10)-2
& log(10)-2 is undefined, hence 'x' is undefined.