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indications equation a^x=x+2

Vali

Member
Dec 29, 2018
48
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!
 

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Last edited:

MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,684
St. Augustine, FL.
I would begin with:

\(\displaystyle f(x)=a^x-x-2\)

Hence, let's examine:

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

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

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

\(\displaystyle f''(x)<0\)

Can you finish?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,643
Leiden
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! ;)
 

Vali

Member
Dec 29, 2018
48
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.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
344
By the way, letting y= x+ 2, the equation becomes

\(\displaystyle a^{y-2}= a^{-2}a^y= a^{-2}e^{ln(a^y)}= a^{-2}e^{yln(a)}= y\).

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

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

Of course, since the W function can be multivalued, that does not answer the original question!