Question #27399

3 Answers
Oct 22, 2017

x = 0 and x = ln(5)

Explanation:

Given two equations:

y=3-e^x" [1]"
y=5(e^-x )-3" [2]"

At the point of intersection y = y, therefore, we can set the right side of equation [2] equal to the right side of equation [1]:

5(e^-x )-3=3-e^x

Add 3 to both sides of the equation:

5(e^-x )=6-e^x

Multiply both sides of the equation by e^-x

5(e^-x )^2=6(e^-x)-1

If we let u = e^-x we can recognize a quadratic equation:

5u^2=6u-1

Write the quadratic so that it is equal to 0:

5u^2-6u+1 = 0

Factor:

(u-1)(5u-1) = 0

u = 1 and u = 1/5

reverse the substitution

e^-x = 1 and e^-x = 1/5

Use the natural logarithm on both sides of both equations:

-x = ln(1) and -x = ln(1/5)

x = 0 and x = ln(5)

Check x = 0:

y=3-e^0
y=5(e^-0 )-3

y=3-1
y=5(1 )-3

y=2
y=2

check x = ln(5):

y=3-e^(ln(5))
y=5(e^-ln(5) )-3

y=3-5
y=5(1/5 )-3

y = -2
y = -2

Both check.

Oct 22, 2017

(0,2), and, (ln5,-2).

Explanation:

To find the point/s of intersection the curves

C_1 : y=3-e^x, and, C_2 : y=5e^-x-3, we need to solve their eqns.

:. 3-e^x=y=5e^-x-3.

:. 3-e^x-5/e^x-3. Let, e^x=t.

:. 3-t=5/t-3 rArr 3t-t^2=5-3t, or, t^2-6t+5=0.

:. (t-5)(t-1)=0.

:. t=5, or, t=1.

t=5 rArr e^x=5 rArr x=ln5. Also, y=3-e^x=3-5=-2.

:. (x,y)=(ln5,-2).

Similarly,

t=1 rArr e^x=1 rArr x=0, and, y=3-e^x=3-1=2.

:. (x,y)=(0,2).

Thus, C_1 nn C_2={(0,2), (ln5,-2)}.

Enjoy Maths.!

Oct 22, 2017

The intersection points will be ( ln5, -2) and (0,2).

Explanation:

This is the explanation for this:

Firstly we equalise both of the equations to find the common x value(s).

5*(e^-x)-3=3-e^x
5*(e^-x)+e^x-3-3=0
5*1/(e^x)+e^x-6=0
5/e^x+e^x-6=0
cancel(e^x)*5/cancel(e^x)+e^x*e^x-6*e^x=0
5+e^(2x)-6*e^x=0
e^(2x)-6*e^x+5=0
(e^x-5)*(e^x-1)=0

We equalise each bracket with 0.
e^x-5=0
e^x=5
lne^x=ln5
xlne=ln5
x*1=ln5
x=ln5 (first x value)

e^x-1=0
e^x=1
lne^x=ln1
x*lne=0
x*1=0
x=0 (second x value)

Then we need to find y-values for each corresponding x value by substituting x value in one of the main equations.
e.g let's substitute each value of x in y=3-e^x

f(ln5)=3-e^ln5
f(ln5)=3-cancele^cancel(ln)5
f(ln5)=3-5
f(ln5)=-2

first intersection point is (ln5,-2).

Let's substitute the other x-value:

f(0)=3-e^0
f(0)=3-1
f(0)=2

second intersection point is (0,2)