Can you solve [tex]2^x = e^{(x+2)}[/tex]?

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)

Answered by AL2006 | 2024-06-10

2 x = e x + 2 l n ( 2 x ) = l n ( e x + 2 ) x l n ( 2 ) = ( x + 2 ) l n ( e ) x l n ( 2 ) = x + 2 x x + 2 ​ = l n ( 2 ) x x ​ + x 2 ​ = l n ( 2 ) 1 + x 2 ​ = l n ( 2 ) x 2 ​ = l n ( 2 ) − 1 x = l n ( 2 ) − 1 2 ​ ​

Answered by Ryan2 | 2024-06-10

To solve the equation 2 x = e ( x + 2 ) , we take the natural logarithm of both sides, rearrange the terms, and isolate x . The solution is found to be approximately x ≈ − 6.518 .
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Answered by AL2006 | 2024-12-23