# Solve for X. e^2x - 3e^x + 2=0

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e^2x-3e^x +2 =0

Let us assume that e^x =y.

Now substitute:

==> y^2 -3y +2 =0

Factorize:\

(y-2)(y-1)=0

y1= 2 ==> y1=e^x1= 2 ==> x1=ln2

y2=1 ==> y2=e^x2= 1 ==> x2=0

The solution for the equation is :

x= {0, ln2}

Being an exponential equation, we'll solve it using the technique of substitution.

We'll substitute e^x by the other unknown, t.

e^x = t

The equation will become:

t^2 - 3t + 2 = 0

We'll apply the quadratic formula:

t1 = [3+sqrt(3^2 - 4*1*2)]/2*1

t1 = (3+1)/2

t1 = 2

t2 = (3-1)/2

t2 = 1

Now, let's find put x:

e^x = t1

e^x = 2

ln e^x = ln 2

x*ln e = ln 2

**x1 = ln 2**, when ln e = 1

e^x = t2

e^x = 1

e^x = e^0

**x2 = 0**

The solutions of the equation are both convenient and they are: {0, ln 2}.

To silve e^2x-3e^2x+2 = 0

Solution:

This is a qudratic in e^x.

So (e^x)^2 -3e^x +2 = 0. By grouping'

e^2x - 2e^x - e^x +2 = 0.

e^x(e^x - 2) -1(e^x - 2) = 0. Pulling out e^x-2, we get:

(e^x - 2)(e^x - 1) = 0. So by zero product rule, we get:

e^x = 2, which gives x = ln2.

e^x = 1, giving lnx = ln1 = 0

The equation e^2x - 3e^x + 2=0 has to be solved. Let e^x = y.

The equation can now be written as:

y^2 - 3y + 2 = 0

Factor the expression on the left

y^2 - 2y - y + 2 = 0

y(y - 2) - 1(y - 2) = 0

(y - 1)(y - 2) = 0

y = 1 and y = 2

Now y = e^x

e^x = 1 gives x = 0

e^x = 2 gives x = ln 2

The solution of the equation e^2x - 3e^x + 2=0 is x = 0 and x = ln 2

e^2x - 3e^x + 2= 0

(e^x)^2 - 3e^x + 2 = 0

(e^x)^2 - 2e^x - e^x + 2 = 0

e^x(e^x - 2) - 1(e^x - 2) = 0

(e^x - 1)(e^x - 2) = 0

e^x = 1 e^x = 2

or e = xth root of1 |or e = xth root of 2

therefore ,

1^(1/x) = 2 ^ (1/x)

Since 1 to the power any number = 1

1 = 2^1/x

we see that only when x = 0 the condition of the equation satisfies

therefore x = 0