Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do I solve this exponential equation?
$$5^{x}-4^{x}=3^{x}-2^{x}$$

share|improve this question
5  
$x=1{}{}{}{}{}$ –  Will Jagy Feb 2 '13 at 4:08
3  
$x=0,1$ seems to be the only real solution. for $x\gt 1$, the L.H.S. growth rate is more than that of R.H.S. and for $x\lt 0$, L.H.S. $\lt $ R.H.S –  Aang Feb 2 '13 at 4:09
    
@WillJagy You forgot $x = 0$ –  hjpotter92 Feb 2 '13 at 4:09
2  
No, it is a religious prohibition. Well, it will be, once I have finished designing the religion. –  Will Jagy Feb 2 '13 at 4:10
1  
@BackinaFlash: Will forgot nothing. :-) –  Brian M. Scott Feb 2 '13 at 4:11

4 Answers 4

Hint:: $\displaystyle \frac{5^x-4^x}{5-4} = \frac{3^x-2^x}{3-2}$

Now Use LMVT (Lagrange mean value theorem) in $(2,3)$ and $(4,5)$

These two are equal only when $x = 0$ and $x=1$

share|improve this answer

$$5^x - 4^x = \int_4^5 x y^{x-1} \,dy$$ $$3^x - 2^x = \int_2^3 x y^{x-1} \,dy$$ $$= \int_4^5 x (y-2)^{x-1} \,dy$$ So the difference between $5^x - 4^x$ and $3^x - 2^x$ is $$ \int_4^5 x (y^{x-1} - (y-2)^{x-1})\,dy$$ In the integrand here, since $y \rightarrow y^{x-1}$ is monotone whenever $x \neq 1$, the expression $(y^{x-1} - (y-2)^{x-1})$ will either be always negative or always positive if $x \neq 1$, in which case the integral itself will be nonzero unless $x = 0$.

We conclude that as long as $x \neq 0$ or $1$, $(5^x - 4^x) - (3^x - 2^x)$ is nonzero. Hence $3^x - 2^x = 5^x - 4^x$ only when $x = 0$ or $1$.

After writing all this out, I probably prefer the mean value theorem approach, but hey it's good to have more than one way of looking at a problem.

share|improve this answer

Only method I can think of is trying graphically. Since their slopes vary a lot after $(1,1)$ and before $(0,0)$, all possible intersections can lie between these two points.

How to find all those possible intersections is something I'd need to work out.

share|improve this answer

Let $f:[0.25,2] \to \mathbb R \,;\, f(a)=(3.5-a)^x+(3.5+a)^x \,.$

Then $f'(a)=x[(3.5+a)^{x-1}-(3.5-a)^{x-1}]$.

Claim: If $x \notin \{ 0,1 \}$ we have $f'(a) \neq 0 \forall a $.

Indeed, in this case $f'(a) =0 \Rightarrow (3.5+a)^{x-1}=(3.5-a)^{x-1} \Rightarrow a=0$ which is not in our domain.

This proves that for $x \neq 0,1$, $f$ is one to one on our domain, and hence

$$f(0.5) \neq f(1.5) \Rightarrow 3^x+4^x \neq 2^x+5^x $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.