Solve the equation for x, accurate to three decimal places: e^4x + 5e^2x = 36.

Question

I really need help

Answer 1

This problem requires a bit of mathematical trickery. This *kind of* looks like a quadratic equation a little bit. Let's try forcing it into that form. The first term, e^4x is the same thing as (e^2x)² so let's write it out like that

 

(e^2x)² + 5(e^2x) - 36 = 0

 

This really looks like a quadratic equation, but with the usual x replaced by e^2x. So let's make the substitution and see what happens. We can replace the e^2x with some other variable V

 

V² + 5V -36 = 0

 

Factor this out to get

 

(V + 9)(V - 4) = 0

 

The roots are -9 and +4

 

Now think about our substitution. Can e^2x ever take on a negative value? No, it can't. So we are left with +4. We can now equate our root from the "quadratic" to what we subbed into the equation. In math-speak, that means

 

4 = e^2x

 

To solve for x, we take the natural log of both sides, noting that the natural log of e^(anything) is defined to be that 'anything'

 

ln(4) = 2x

 

ln(4)/2 = x

 

Plug this into a calculator to get

 

x = 0.693147

 

to three decimal places, that is

 

x = 0.693

 

I hope this helps!

Answer 2

e^4x+5e^2x=36, solve for x.

 

Let u = e^2x, then the following equation is true.

 

u²+5u=36, subtract 36 on both sides of the equation.

 

u²+5u-36=0, factor the equation.

 

(u+9)(u-4)=0, therefore,

 

u+9=0 and/or u-4=0, solve for u.

 

u=-9 and/or u=4; replaced the real value of u back into the answers.

 

e^2x=-9 and/or e^2x=4; do the inverse of e which is natural log.

 

ln(e^2x)=ln(-9)  and/or  ln(e^2x)=ln(4); since the natural log cannot have a negative number then you are left with,

 

ln(e^2x)=ln(4); simplify

 

2x=ln(4); divide by 2 of both sides of the equation.

 

x=^ln(4)/₂ = 0.693

Answer 3

Violet,

 

The key here is to rewrite this as a quadratic equation as follows:

W^2+5W-36=0 where W=e^2x

Which can be factored or solved using the quadratic formula into

(W+9)(W-4)=0

 

Now set (W+9) and (W-4) to zero. Then W=-9 or W=4.

 

Remember that W is an something raised to a exponent which is always greater than zero. This means we can assume in the case of (W+9), there is no real solution. However, when (W-4)=0, e^2x=(e^x)^2=4, making e^x plus or minus two. Since something raised to a exponent cannot be negative, e^x=2.

 

To solve for x, we use something called the natural log or ln. It works like this: e^1=e so ln e=1. With natural logs, the base of the exponent is assumed to be e. The final result is the base raised to the exponent and the exponent is 1. Natural logs find the exponent that e can be raised to find a number. For example, ln 1 is zero because anything to the zeroth power is one. 

 

In this case e^x=2, so x=ln 2, which is a number you can punch in your calculator. As for e, it's a constant someone defined. Go ahead and punch in e^1 in your calculator if you're curious. 

 

Hope that helps!