how to solve this equation? HELP!!!?

Question

This is a question is under logarithms.

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

Pls help..

Answer 1

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 2

Is question meant to read 2e^(x) = 3 - e^(x - 1) ?

If so :-
2e^(x) + e^(x - 1) = 3
e^(x).[ 2 + e^(-1)] = 3
e^(x) (2 . 37) = 3
e^(x) = 1.27
x = log (1.27)
x = 0.24