MHB What are the positive values for a in the equation a^x=x+2?

  • Thread starter Thread starter Vali
  • Start date Start date
AI Thread Summary
The equation a^x = x + 2 has two real solutions, and the discussion focuses on identifying the positive values for "a" that allow this. It is established that for a > 1, the function decreases from infinity to -1 and then increases, leading to two intersection points with the line y = x + 2. For values of a between 0 and 1, only one intersection point exists due to the opposite slopes of the graphs. The second derivative test is discussed, indicating that for a to yield two roots, it must be greater than 1. The use of Lambert's W function is also mentioned as a method to explore solutions further.
Vali
Messages
48
Reaction score
0
Sorry for posting again but I need to prepare for exam.
a^x=x+2 has two real solutions.I need to find positive values for "a".
A) (1, infinity)
B) (0,1)
C) (1/e , e)
D) (1/(e^e), e^e)
E) (e^(1/e), infinity)
I tried to solve and I did it but I don't understand some things.
I let a picture below to see.
First, I need to know if there's other way to solve this kind of exercise.I would be happy if I would get some ideas.
Also, from my solution, I don't understand why from that table results just one solution and from the graphic results two solutions.Usually, to see the number of solutions I use this kind of table.
For a>1 f decrease from infinity to -1, then increase from -1 to infinity.I'm really confused.Need some indications here.
Thank you!
 

Attachments

  • expo.jpg
    expo.jpg
    42.5 KB · Views: 109
Last edited:
Mathematics news on Phys.org
I would begin with:

$$f(x)=a^x-x-2$$

Hence, let's examine:

$$f'(x)=a^x\ln(a)-1$$

$$f''(x)=a^x\ln^2(a)$$

Now, in order for \(f(x)\) to have 2 roots, we require upward concavity, that is we require:

$$f''(x)<0$$

Can you finish?
 
Hi Vali,

You have a mistake for $a<1$ where you took the example $a=-e$.
Admittedly $-e < 1$, but $a^x$ is not defined for negative $a$.
So we should pick $0<a<1$. We can pick for instance $a=\frac 1e$ so that we get $a^x=(\frac 1e)^x = e^{-x}$.

To understand better what's going on, let's draw a couple of graphs.

\begin{tikzpicture}[scale=0.6]
\begin{scope}
\draw[help lines] (-4,-2) grid (4,5);
\draw[<->] (-4.4,0) -- (4.4,0) node
{$x$};
\draw[<->] (0,-2.2) -- (0,5.2) node[above] {$y$};
\draw foreach \i in {-4,-3,-2,-1,1,2,3,4} { (\i,0.1) -- (\i,-0.1) node[below] {$\i$} };
\draw foreach \i in {-2,-1,1,2,3,4,5} { (0.1,\i) -- (-0.1,\i) node
{$\i$} };
\draw[domain=-4:2.2, variable=\x, red, ultra thick] plot ({\x}, {(\x+2)}) node
{$y=x+2$};
\draw[domain=-4:1.6, variable=\x, blue, ultra thick] plot ({\x}, {exp(\x)}) node
{$y=a^x, a>1$};
\end{scope}
\begin{scope}[xshift=10cm]
\draw[help lines] (-4,-2) grid (4,5);
\draw[<->] (-4.4,0) -- (4.4,0) node
{$x$};
\draw[<->] (0,-2.2) -- (0,5.2) node[above] {$y$};
\draw foreach \i in {-4,-3,-2,-1,1,2,3,4} { (\i,0.1) -- (\i,-0.1) node[below] {$\i$} };
\draw foreach \i in {-2,-1,1,2,3,4,5} { (0.1,\i) -- (-0.1,\i) node
{$\i$} };
\draw[domain=-4:2.2, variable=\x, red, ultra thick] plot ({\x}, {(\x+2)}) node[above] {$y=x+2$};
\draw[domain=-4:3, variable=\x, blue, ultra thick] plot ({\x}, {1}) node[above] {$y=a^x, a=1$};
\end{scope}
\begin{scope}[xshift=20cm]
\draw[help lines] (-4,-2) grid (4,5);
\draw[<->] (-4.4,0) -- (4.4,0) node
{$x$};
\draw[<->] (0,-2.2) -- (0,5.2) node[above] {$y$};
\draw foreach \i in {-4,-3,-2,-1,1,2,3,4} { (\i,0.1) -- (\i,-0.1) node[below] {$\i$} };
\draw foreach \i in {-2,-1,1,2,3,4,5} { (0.1,\i) -- (-0.1,\i) node
{$\i$} };
\draw[domain=-4:2.2, variable=\x, red, ultra thick] plot ({\x}, {(\x+2)}) node
{$y=x+2$};
\draw[domain=-1.6:3, variable=\x, blue, ultra thick] plot ({\x}, {exp(-\x)}) node[above right] {$y=a^x, 0<a<1$};
\end{scope}
\end{tikzpicture}

Note that the line $y=x+2$ intersects the y-axis at $2$, which is always above the $1$ where $y=a^x$ intersects the y-axis.
The cases are:
  • For $a>1$, the graph of $y=a^x$ always slopes upwards exponentially so that it will always overtake the line.
    In other words, we always have 2 intersection points.
  • For $a=1$ we have indeed always 1 intersection point as we have 2 intersecting lines.
  • For $0<a<1$ we also have always 1 intersection point since the graphs have opposite slopes.
    As you can see, this is different from what you had, as $a$ must be positive for a proper definition of $a^x$.
  • For $a=0$ (not drawn) $a^x$ is only defined for positive $x$ where it is $0$, so no intersection points.
    We can't divide by $0$ after all.
  • For $a<0$ the expression $a^x$ is undefined for real $x$. We can't take roots (at e.g. $x=\frac 12$) of negative numbers after all.
I hope this clarifies a bit and gives you a different way to look at the problem! ;)​
 
Thank you for your help!I understood.

I try to understand the method with second derivative.
f''(x) < 0 has no solution for x real.
 
By the way, letting y= x+ 2, the equation becomes

[math]a^{y-2}= a^{-2}a^y= a^{-2}e^{ln(a^y)}= a^{-2}e^{yln(a)}= y[/math].

Letting z= y ln(a), a^{-2}e^{z}= \frac{z}{ln(a)}. ze^{-z}= a^{-2}ln(a). Finally, letting u= -z, ue^u= -a^{-2}ln(a). Then u= W(-a^{-2}ln(a)) where W is "Lambert's W function", the inverse to f(x)= xe^x.

Then z= -W(-a^{-2}ln(a)), y= -\frac{W(-a^{-2}ln(a))}{ln(a)}, x= -\frac{W(-a^{-2}ln(a))}{ln(a)}- 2.

Of course, since the W function can be multivalued, that does not answer the original question!
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...

Similar threads

Back
Top