Solve x when it is an exponent... 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.
 A: 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. 
A: 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.
A: Hint:
$$(5.5)^{2x} = (11/2)^{2x} = (121/4)^x$$
and
$$(6.5)^x = (13/2)^x$$
Can you take it from here?
A: 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 $\mathbb{R}$.
