Question

Question #27399

Answer 1

`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.

Answer 2

`(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.!

Answer 3

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)`