Difficult: For which a>0 does y=a^x intersect y=x? etc

[dts5044]

1. For which a > 0 does the curve y = a[sup:329hiy5d]x[/sup:329hiy5d] intersect the line y = x?

2. For which a > 0 is it true that a[sup:329hiy5d]x[/sup:329hiy5d] >= x + 1 for all real numbers x?

3. For what values of k does the equation e[sup:329hiy5d]2x[/sup:329hiy5d] = k * (x)[sup:329hiy5d]1/2[/sup:329hiy5d]

I have worked for hours on each problem. My teacher gave me the hint of using the equation f(x) = a[sup:329hiy5d]x[/sup:329hiy5d] - x for the first one, and using its derivative to solve the problem. I always seem to run into trouble with the fact that both a and x can change..

any help is greatly appreciated. Thanks!

 

[galactus]

#1:

I will go ahead and show you one way of doing it. Perhaps that will give you an idea on the others.

\(\displaystyle \frac{d}{dx}[a^{x}]=ln(a)\cdot{a^{x}}\)

Since the line it intersects has slope 1 everywhere, then the derivative must be 1 at some point x.

\(\displaystyle ln(a)\cdot{a^{x}}=1\)

\(\displaystyle x=\frac{ln(\frac{1}{ln(a)})}{ln(a)}\)

Let's sub this in \(\displaystyle a^{x}=x\) and solve for a:

\(\displaystyle a^{\frac{ln(\frac{1}{ln(a)})}{ln(a)}}=\frac{ln(\frac{1}{ln(a)})}{ln(a)}\)

This looks worse than it is.

Rewrite as:

\(\displaystyle e^{\frac{1}{ln(a)}}=(\frac{1}{ln(a)})^{\frac{1}{ln(a)}}\)

Take ln to both sides and get:

\(\displaystyle \frac{1}{ln(a)}=\frac{1}{ln(a)}\cdot{ln(\frac{1}{ln(a)})}\)

Divide by 1/ln(a):

\(\displaystyle 1=ln(\frac{1}{ln(a)})\)

Take e to both sides:

\(\displaystyle e=\frac{1}{ln(a)}\)

\(\displaystyle \frac{1}{e}=ln(a)\)

\(\displaystyle \boxed{e^{\frac{1}{e}}=a}\)

There may be easier ways, but this helps you to see how to manipulate a log if need be.

Now, you try the others.

 

[wjm11]

1. For which a > 0 does the curve y = a^x intersect the line y = x?

Now that Galactus has done all the "heavy lifting," I'll suggest the trivial solution: a = 1 (at y = x = 1).

 

[skeeter]

a[sup:2bvw7awo]x[/sup:2bvw7awo] = x , a > 0

a[sup:2bvw7awo]1/4[/sup:2bvw7awo] = 1/4 ... a = 1/4[sup:2bvw7awo]4[/sup:2bvw7awo]

a[sup:2bvw7awo]1/3[/sup:2bvw7awo] = 1/3 ... a = (1/3)[sup:2bvw7awo]3[/sup:2bvw7awo]

a[sup:2bvw7awo]1/2[/sup:2bvw7awo] = 1/2 ... a = (1/2)[sup:2bvw7awo]2[/sup:2bvw7awo]

a[sup:2bvw7awo]3/4[/sup:2bvw7awo] = 3/4 ... a = (3/4)[sup:2bvw7awo]4/3[/sup:2bvw7awo]

a[sup:2bvw7awo]3/2[/sup:2bvw7awo] = 3/2 ... a = (3/2)[sup:2bvw7awo]2/3[/sup:2bvw7awo]

a[sup:2bvw7awo]2[/sup:2bvw7awo] = 2 ... a = 2[sup:2bvw7awo]1/2[/sup:2bvw7awo]

a[sup:2bvw7awo]3[/sup:2bvw7awo] = 3 ... a = 3[sup:2bvw7awo]1/3[/sup:2bvw7awo]

a[sup:2bvw7awo]4[/sup:2bvw7awo] = 4 ... a = 4[sup:2bvw7awo]1/4[/sup:2bvw7awo]

see a pattern?