The diagram shows the curves y = e^2x−3 and y = 2 ln x. When x = a the tangents to the curves are parallel. The diagram basically shows two curve lines not touching each other.

Show that a satisfies the equation a= 1/2 (3-ln a)

Verify by calculation that this equation has a root between 1 and 2

can someone help me please!!! Thanks in advance.

Question

The diagram shows the curves y = e^2x−3 and y = 2 ln x. When x = a the tangents to the curves are parallel. The diagram basically shows two curve lines not touching each other. 

Show that a satisfies the equation a= 1/2 (3-ln a) 

Verify by calculation that this equation has a root between 1 and 2 

can someone help me please!!! Thanks in advance.

nincompoop · Answer

WHERE  IS THE DIAGRAM?

anonymous · Answer

there you go please help me I'm dying trying to solve this question :3

mathmale · Answer

Navina:  Your diagram doesn't match your typewritten problem statement.  On which question shall we focus right now?

anonymous · Answer

sorry I send the wrong question and diagram. Here you go.

mathmale · Answer

OK.  Sure this is the problem you want to discuss?  If so:  How do you find the SLOPE of the tangent line to such a curve at any x-value, 'a'?

mathmale · Answer

What is the slope of the tangent line to the curve $$y=e ^{2x-3}$$ at x=a?

anonymous · Answer

using the formula Y=mX + c ? I make both equations into one equation but I got stuck some where at there

mathmale · Answer

Let's go back to basics.  Why do we bother to learn the concept of "derivative" in calculus?  What's the significance of the "derivative?"

anonymous · Answer

finding the slope of a tangent line?

mathmale · Answer

Yes.  Among other things, the derivative of a function, evaluated at the x value 'a,'  represents the slope of the tangent line to the graph of that function at x=a.

mathmale · Answer

So again I ask you:  What is the derivative of $$y=e ^{2x-3} $$

mathmale · Answer

And what is the derivative of y=2ln x?

anonymous · Answer

$$2e ^{2x-3}$$

mathmale · Answer

That's the derivative of the first function.  Very good.

anonymous · Answer

and $$\frac{ 2 }{ x }$$

anonymous · Answer

sorry $$\ln \frac{ 2 }{ x }$$

mathmale · Answer

Why the 'ln'?

mathmale · Answer

What's the derivative of y=ln x?

anonymous · Answer

owh sorry again it's $$\frac{ 2 }{ x}$$

mathmale · Answer

Now the problem statement says that the two separate tangent lines are parallel.

What does that tell you about the quantities $$\frac{ d }{ dx }e ^{2x-3}~and~\frac{ d }{ dx }2*\ln x?$$

anonymous · Answer

both have the same gradient

mathmale · Answer

No.  These two quantities (these two derivatives) are EQUAL.

mathmale · Answer

Write an equation involving the two derivatives that you've already found.

anonymous · Answer

I did that and ln both equations both I got stuck there

mathmale · Answer

Please write out the equation.

anonymous · Answer

$$2e ^{2x-3} = \frac{ 2 }{ x }$$

mathmale · Answer

Good.  Now divide both sides by 2 to simplify the equation.

anonymous · Answer

$$e^{2x-3}= \frac{ 1 }{ x}$$

mathmale · Answer

Good.  
The first two methods that come to mind in regard to solving this equation are:
1) Graphing the function on the left side of this equation AND the function on the right side of this equation on the same set of coordinate axes, for x>0, and estimating (by eye) the approx. solution (that is, the x-coordinate of the intersection of the two curves)., and
2) Using Newton's Method to determine the solution to the desired number of decimal places.
Your choice?

anonymous · Answer

Can we ln both side? is it possible?

mathmale · Answer

As I said, the two methods I've listed were the first that came to mind.  Of course you can take the natural log (or any other log) of both sides of this equation.  Perhaps that will simplify the solution.  Again:  Your choice.

anonymous · Answer

Okay then I will go with the Log method

mathmale · Answer

Good.  Take the natural log of both sides of your equation:

anonymous · Answer

$$\ln 2e ^{2x-3}= \ln \frac{ 2 }{ x }$$

mathmale · Answer

that's OK, but remember that you and I have already eliminated the '2' factor from both sides of this equation.  
Go ahead and evaluate the natural log of each side of the equation, separately.

mathmale · Answer

sorry for the delay. OpenStudy was temporarily hung up ("Lost the connection")

anonymous · Answer

$$(2x-3)\ln2e = \ln 2-\ln x$$

anonymous · Answer

It's okay :)

mathmale · Answer

Or...$$(2x-3)\ln e = \ln 1 - \ln x$$

mathmale · Answer

Simplify this, remembering that your end goal is to solve for x.

anonymous · Answer

how did you get (2x-3)ln e =ln1 +lnx ?

mathmale · Answer

By taking the natural log of both sides of$$e^{2x-3}= \frac{ 1 }{ x}$$....remember, we can eliminate that factor 2 from both sides simply by dividing both sides by 2.

mathmale · Answer

Note:  It's (2x-3)*ln e = ln 1 - ln x.  Simplify this.

anonymous · Answer

ln e = 1 and ln 1= 0 so, (2x-3) = -ln x

mathmale · Answer

Really good.  Now please estimate x; in other words, attempt an approximate solution.  You could either graph 2x-3 and -ln x on the same set of axes and eyeball the x-coordinate of the point of intersection of these two curves, or you could use Newton's Method to determine the solution more accurately.

anonymous · Answer

I will go with Newton's method

mathmale · Answer

Fine.  Create the function f(x)=2x-3-(-lnx), or f(x)=2x-3+ln x, and proceed from there.  What is f '(x)?

anonymous · Answer

f ' (x) =2+1/x

mathmale · Answer

Good.  Use f(x) and f '(x) in the formula for Newton's Method.

anonymous · Answer

to be honest I didn't learn Newton's Method is there other way to do? sorry for the trouble

anonymous · Answer

Thank you I found the answer already. Thanks for your help wish you were my maths teacher. Have a great day :)

mathmale · Answer

@navina, I'd be delighted to work with you further.  If you want my attention, just tag me when you post future questions.