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Sagot :
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)
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)
[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]
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