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Please help me solve: $$5.5^{2x}-6.5^x +1 = 0$$

I really don't think it is possible at all. They wanted me to do it without an calculator.

By the way I am new to the forum so I don't know if I have broken the rules.

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  • $\begingroup$ Are you sure it isn't $6.5^{2x}-6.5^x +1 = 0$ or $5.5^{2x}-5.5^x +1 = 0$? Perhaps $5·5^{2x}-6·5^x +1 = 0$? These three are solvable if you consider it a second-degree equation on $5^x$ or some other exponential function of $x$. The solutions of the last are $x=1$ and $x=0 $ $\endgroup$ Commented Sep 25, 2015 at 21:23
  • $\begingroup$ Neither of those above, I really doubt this question is possible... it was set as an A-level question according to my friend... $\endgroup$ Commented Sep 25, 2015 at 21:27
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    $\begingroup$ This has no real solutions. $\endgroup$ Commented Sep 25, 2015 at 21:39

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One has to compare $f(x) = 5.5^{2x}$ with $g(x) = 6.5^x$. $f$ can be transformed into $f(x) = (5.5^2)^x$.

So we have a case of $a^x$ vs $b^x$ with $a > b$ and $a, b > 0$.

A solution exists, if $a^x - b^x = -1$

For $x > 0$ we have $$ a > b \Rightarrow a^x > b^x \Rightarrow a^x - b^x > 0 > -1 $$ thus no solution exists.

For $x = 0$ we have $$ a^x - b^x = 1 - 1 = 0 > -1 $$ which is no solution as well.

For $x < 0$ we have $$ a^x = a^{-\lvert x \rvert} = 1 / a^{\lvert x \rvert} \in (0, 1) $$ and $$ b^x \in (0, 1) $$ For a solution $x$ we would need $$ b^x = a^x + 1 $$ which is not possible because any two points $u, v \in (0,1)$ are a distance $d < 1$ apart.

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Hint: $$(5.5)^{2x} = (11/2)^{2x} = (121/4)^x$$ and $$(6.5)^x = (13/2)^x$$ Can you take it from here?

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  • $\begingroup$ What exactly is this hint suggesting? $\endgroup$ Commented Sep 25, 2015 at 22:02
  • $\begingroup$ Well, the equation can then be rewritten as: $$(121)^x - (26)^x + 4^x = 0.$$ The equation has no solutions for the case $x \geq 0$. It thus remains to consider the case $x < 0$. $\endgroup$ Commented Sep 25, 2015 at 22:08
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Since $5.5^{2x}>6.5^x$, you have that $5.5^{2x}-6.5^x+1>0$. Edit: This only shows there are no positive solutions.

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  • $\begingroup$ This is not true for $x < 0$ $\endgroup$ Commented Sep 25, 2015 at 21:42
  • $\begingroup$ @rightskewed You're absolutely right, thanks for correcting me. $\endgroup$ Commented Sep 25, 2015 at 21:58
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Consider the function $$f:\mathbb{R}\to\mathbb{R}$$$$x\mapsto30.25^x-6.5^x+1$$ Because $\forall x\in\mathbb{R},\,5.5^{2x}=(5.5^2)^x=30.25^x$ your equation in $\mathbb{R}$ is equivalent to the equation $f(x)=0$ in $\mathbb{R}$.

For simplification, let $a=30.25$ and $b=6.5$. Then $\forall x\in\mathbb{R},\,f(x)=a^x-b^x+1$. This function is differentiable on $\mathbb{R}$ and we have $\forall x\in\mathbb{R},\,f'(x)=(\ln a)a^x-(\ln b)b^x$.

We're going to study the variations of $f$. Let $x\in\mathbb{R}$ such as $f'(x)=0$. Thus:$$(\ln a)a^x-(\ln b)b^x=0$$$$(\ln a)a^x=(\ln b)b^x$$$$\ln\ln a +x\ln a=\ln\ln b+x\ln b$$$$(\ln a-\ln b)x=\ln\ln b -\ln\ln a$$$$x=-\dfrac{\ln\ln a -\ln\ln b}{\ln a-\ln b}$$

Let $x_0=-\dfrac{\ln\ln a -\ln\ln b}{\ln a-\ln b}$.In the equalities above, if you replace the relation $=$ by $>$ and $<$ ($a>b\Rightarrow \ln a-\ln b>0$) you can conclude that $f'$ is positive on $(x_0,+\infty)$, negative on $(-\infty,x_0)$ and $f'(x_0)=0$. Thus $f(x_0)$ is the minimum value of $f$ on $\mathbb{R}$, i.e, $\forall x\in\mathbb{R},\,f(x)\ge f(x_0)$. Using a calculator or a software (personally I used Maple), we find that $f(x_0)\approx0.7826480697$. Therefore your equation has no solution on $\mathbb{R}$.

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    $\begingroup$ Clearly the minimum of $f$ is less than $f(0) = 1$. $\endgroup$ Commented Sep 25, 2015 at 22:09
  • $\begingroup$ @mrf Thanks for noticing this to me. My bad I calculated $f(-x_0)$ instead of $f(x_0)$. $\endgroup$ Commented Sep 25, 2015 at 22:12
  • $\begingroup$ I am really in over my head but this answer looks really really cool. Hopefully by the end of my further maths course, I can decipher half of this. By the way thanks to everyone for replying :)! $\endgroup$ Commented Sep 26, 2015 at 9:13
  • $\begingroup$ @ChrDueSukPe So I guess you're a freshman at highschool. Indeed you'll be able to understand my answer if you become a junior highschool student; at that year you will study differentiation and the exponential and logarithm functions. Anyway, I appreciate you like my answer, thank you! $\endgroup$ Commented Sep 26, 2015 at 9:17
  • $\begingroup$ @ChrDueSukPe By the way since you understand mvw's answer (that must be the case!) you should accept his answer ;) $\endgroup$ Commented Sep 26, 2015 at 9:20

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