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I have problem with equation: $4^x-3^x=1$.

So at once we can notice that $x=1$ is a solution to our equation. But is it the only solution to this problem? How to show that there aren't any other solutions?

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  • $\begingroup$ You may re-write your equation as \begin{align} 1&=4^x-3^x\\ &=e^{x\ln{4}}-e^{x\ln{3}}\\ \end{align} Maybe that will help get us going $\endgroup$ – Logan Tatham Feb 10 '14 at 21:29
  • $\begingroup$ I recommend using Rolle's theorem. $\endgroup$ – IAmNoOne Feb 10 '14 at 21:31
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Let $f(x)=1-\left({3\over4}\right)^x-\left({1\over4}\right)^x$. The functions $(3/4)^x$ and $(1/4)^x$ are strictly decreasing functions of $x$, so the minus signs make $f(x)$ strictly increasing.

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  • $\begingroup$ I don't understand how do you obtain $(\frac{3}{4})^x$ and $(\frac{1}{4})^x$ could you explain it? $\endgroup$ – Jessica Feb 10 '14 at 21:49
  • $\begingroup$ @Jessica, I divided both sides of your equation by $4^x$ and then moved everything over to the left hand side. $\endgroup$ – Barry Cipra Feb 10 '14 at 21:50
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Prove it by contradiction.

Let $f(x) = 4^x - 3^x - 1$. This function is smooth and as you have shown, $x = 1$ is a root. By Rolle's theorem, suppose it had two other roots $f(a) = f(b) = 0$, then there exists $c$ in between $a$ and $b$ (where $a < 1 < b$) such that $f'(c) = 0$, but notice for $ x \geq 0$

$$f'(x) = 4^x \ln 4 - 3^x \ln 3 \geq 3^x (\ln 4 - \ln 3) > 0.$$

and for $x < 0$, we have

$$f'(x) = 4^x \ln 4 - 3^x \ln 3 \leq \ln 4 (4^x - 3^x) < 0.$$

Hence you can conclude there is only one root.

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Hint: one way of showing that a function takes a value only once is to show that it is increasing.

Hint: For negative values of $x$ you need a different observation.

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  • $\begingroup$ I thought to rewrite equation as $4^x=3^x+1$ and for $x \le 0$ we have that $4^x \in (0,1]$, and $(3^x+1) \in (1,2]$ so no solutions here for $x>1$ we have $(3+1)^x=3^x+1$ and I don't know how to prove that $(3+1)^x>3^x+1$, and I haven't finised for $x \in (0,1)$ yet, but what do you think about it? $\endgroup$ – Jessica Feb 10 '14 at 21:43
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You can make a graph of the function $y=4^x-3^x-1=0$. It has a y-intercept at (0,-1) and for $x<0$ the curve stays under the y-axis and for $X>0$ the curve is only increasing.

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Show that $4^x$ grows much faster than $3^x$.

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    $\begingroup$ It is not often that a one-liner makes a good Answer. There's a lot that needs to be done to flesh out your idea into a workable approach. E.g. $4^x$ grows much faster than $3x$, but there's more than one solution to $4^x - 3x = 1$. $\endgroup$ – hardmath Feb 11 '14 at 1:36
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It is easy to see that there is no negative solution.

Assume by contradiction there exists a solution $x=a$ with $0 < a \neq 1$.

Let $g(y)=y^a$. By the mean value Theorem, there exists some $c \in (3,4)$ such that

$$\frac{g(4)-g(3)}{4-3} =g'(c)$$

Therefore $$1= g'(c)=ac^{a-1}$$

But this is impossible: if $a>1$ then $a >1$ and $c^{a-1}>1$, while if $a<1$ then $0< a<1$ and $0< c^{a-1} <1$.

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