Respuesta :

AL2006
Answer: Yes, I can.


Although you haven't asked for the solution, here it is anyway:

2^x = e^(x+2)

x ln(2) = x+2

x ln(2) - x = 2

x [ ln(2) - 1 ] = 2

x = 2 / [ ln(2) - 1 ]

x = 2 / -0.3069... = - 6.518... (rounded) 

Ryan2
[tex]2^x=e^{x+2}\\ \\ ln(2^x)=ln(e^{x+2})\\ \\ xln(2)=(x+2)ln(e)\\ \\ xln(2)=x+2\\ \\ \frac{x+2}{x}=ln(2)\\ \\ \frac{x}{x}+\frac{2}{x}=ln(2)\\ \\ 1+\frac{2}{x}=ln(2)\\ \\ \frac{2}{x}=ln(2)-1\\ \\ \boxed{x=\frac{2}{ln(2)-1}}[/tex]