I do not know how to solve for x!?

Question

Please help:

4x^2 * 3^x + 2x * 3^x

How do I solve for x? Please show steps. Finally, how do I prove this is true?

e^(-x) * e^(x+1) = e

I meant to write equal to 0

Answer 1

(4x^2) * (3^x) + (2x )* (3^x) = 0

factor out 3^x : (3^x)*[ (4x^2) + (2x) ] = 0
and we got 3^x = 0, since we can't solve x for this, so just leave it.

the remaining (4x^2) + (2x) = 0
factor out x : x [ 4x + 2] = 0
and we got x = 0

left 4x + 2 = 0
and x = -0.5

therefore we have 3^x = 0, x = 0 and x = -0.5

Second question:
e^(-x) * e^(x+1)

= e^(-x + x + 1)
x cancelled out

= e^1

=e (proven)

Answer 2

I assume you mean:
4x^2 * 3^x + 2x * 3^x= 0. If so, then
3^x(2x)(2x+1)=0
2x+1 = 0 --> x=-1/2
2x=0 --> x = 0
3^x =0 does not have a solution but has a limit of 0 as x approaches - infinity

e^(-x) * e^(x+1) = e
1/e^x * e^x * e^1 = e

Answer 3

For the first question, it can't be solved because there's no equal on it. That's just an answer. If you know how much is the answer, certain I can solve what's x.

For the second:

e^(-x) * e^(x+1) = e

If the based number is same (e), we can add (-x) with (x+1)

So, the solution :

e^ (-x+x+1) = e

e^ (0+1) = e

e^1 = e

It has been proved.