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}
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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
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
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