how to solve this equation? Help!?

Question

use any suitable substitutions to solve this eqn:

2e^x = 3 - e^(x+1)

btw this is under logarithm

Answer 1

x=log y then
2y = 3 - ey
y = 3/(2+e)
x = log [3/(2+e)]

e is approximately 2.718..something

Answer 2

2e^x = 3 - e^(x + 1)

First, move everything with an e to the left hand side.

2e^x + e^(x + 1) = 3

Now, express e^(x + 1) as e * e^x.

2e^x + e(e^x) = 3

Factor e^x,

e^x [2 + e] = 3

Divide [2 + e] both sides,

e^x = 3/[2 + e]

Now, convert this to logarithmic form. Note that
b^c = a in logarithmic form is log[base b](a) = c

log[base e]][3/(2 + e)] = x

But log[base e] is defined to be the natural log, ln.

x = ln(3 / (2 + e))

That is in its exact form, and if you wanted to, you can calculate an approximation with your calculator, by doing the following:

Use the log property that allows you to decompose this as a difference.

x = ln(3) - ln(2 + e)

And solve for that.

Answer 3

2e^x=3-e^(x+1)
=>2e^x=3-e^x*e [e^(x+1)can also be written as e^x*e]
=>2e^x+e^x*e =3
=>e^x(2+e)=3
=>e^x=3/(2+e)
=>x=ln{3/(2+e)} [y=b^x can also be written as x=log(b-base)y]

Therefore, x= -0.452832425

Answer 4

write e^(x+1) = e x e^x:

2e^x = 3 - e x e^x
2e^x + e x e^x = 3
(e^x)(2+e)=3
e^x = 3/(2+e)
x = ln (3/2+e)
x = ln 3 - ln (2 + e)
x = -0.452832425

Answer 5

write e^(x+1) = e*e^x:

2e^x = 3 - e*e^x
(e^x)*(2+e)=3
x= log[3/(2+e)]
= -0.452832425