logs & algebra (1 Viewer)

coyazayo · Mar 27, 2016

Hi to everyone reading this thread. I need help with these questions. Like always thankssss

1. Determine the number of solutions to the equation: (1/2)^(x-1) -1 -log1/2(x)=0

Prove that the solution to the equation is between x=14/10 and x=15/10

2.If x^2-x-1=0, find the exact value of x^12 (I am confused with how you approach these type of questions. Are there any easy way of looking at these types of questions?)

3.Given that x^2+x+1=0, where c is the positive intergers, find the value of P=x^n+1/x^2n

KingOfActing · Mar 27, 2016

coyazayo said:
Hi to everyone reading this thread. I need help with these questions. Like always thankssss

1. Determine the number of solutions to the equation: (1/2)^(x-1) -1 -log1/2(x)=0

Prove that the solution to the equation is between x=14/10 and x=15/10

2.If x^2-x-1=0, find the exact value of x^12 (I am confused with how you approach these type of questions. Are there any easy way of looking at these types of questions?)

3.Given that x^2+x+1=0, where c is the positive intergers, find the value of P=x^n+1/x^2n

$$For question 1, rewrite the equation as $ \left(\frac{1}{2}\right)^{x-1} = 1 + \log_{\frac{1}{2}}x\\$Plotting the two graphs reveals cleary that there is only one solution. $ (x = 1)$

KingOfActing · Mar 27, 2016

InteGrand · Mar 27, 2016

$\noindent In fact, Question 2 is related to the Fibonacci sequence (not surprising, as the quadratic equation is the defining equation for the Golden Ratio $\varphi$, which crops up in many things related to the Fibonacci numbers, among other things). The Fibonacci numbers may be defined as follows: they are the numbers satisfying the recurrence $F_n = F_{n-1}+F_{n-2}$, with $F_0 = 0$ and $F_1 = 1$. So the Fibonacci sequence is $0,1,1,2,3,5,7,\ldots$.$

$\noindent Now, the given equation is $x^2 - x- 1 = 0$, so $x^2 = x+1$ (note that $x\neq 0$). It is easy to show by induction that $x^n $ can be expressed in the form $ax+b$ for any positive integer $n$. I'll give a sketch of this proof here. The case for $n=1$ is trivial, and so is $n=2$, since $x^2=x+1$ from the given equation, so assume that $x^n = ax+b$ for some particular integer $n\geq 1$. Then $x^{n+1}=x\cdot x^n = x(ax+b) = ax^2 + bx $, which is a linear expression in $x$ because $x$ and $x^2$ are linear expressions in $x$, as mentioned in the base case step. The result follows by induction.$

$\noindent Now, we'll find a recurrence relation for these coefficients in the linear expressions. Let $x^n = a_n x + b_n$, for integers $n\geq 1$. Note that since $x=1x+0$ and $x^2 = x+1$, we have $a_1 = 1 = F_1, b_1 = 0 = F_0$, and $a_2 = 1 = F_2, b_2 = 1 = F_1$. Then$

$\begin{align*}x^{n+1} &= x \cdot x^n \\ &= x \left(a_n x + b_n \right) \\ &= a_n x^2 + b_n x \\ &= a_n \left(x+1\right) + b_n x \\ &= \underbrace{\left(a_n + b_n \right)}_{a_{n+1}}x + \underbrace{a_n}_{b_{n+1}}.\end{align*}$

$\noindent So we clearly have: $b_{n+1} = a_n$ and $a_{n+1}=a_n + b_n$, for $n\geq 1$. So for $n\geq 2$, we have $a_{n+1}=a_n + b_n = a_{n-1}+a_n$ (since $b_n = a_{n-1}$ for $n\geq 2$, using what we just found). So since $a_{n+1}=a_{n-1}+a_{n}$, and $a_1 =1=F_1$, $a_2 = 1=F_2$, the $a_n$'s are just the Fibonacci sequence, so $a_n= F_n$. Also, $b_n = a_{n-1}=F_{n-1}$.$

$\noindent Thus $x^{12}=a_{12}x+b_{12}=\color{blue}F_{12} \color{black}x+ \color{red}F_{11} \color{black}$ (in general, $x^n= F_n x + F_{n-1}$, for any integer $n\geq 1$). Now we just need to calculate the right Fibonacci numbers.$

$\noindent The Fibonacci numbers are (starting from $F_1$): $1,1,2,3,5,8,13,21,34,55,\color{red} 89\color{black},\color{blue}144\color{black},...$. So $F_{12}= 144$ and $F_{11}=89$. Hence $x^{12}=144 x + 89$, where $x$ is a solution to the given quadratic, so $x = \varphi = \frac{1+\sqrt{5}}{2}$ or $x=-\frac{1}{\varphi}=\frac{1-\sqrt{5}}{2}$. You can easily simplify this expression of $144 x + 89$ if you wish.$

logs & algebra (1 Viewer)

coyazayo

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KingOfActing

lukewarm mess

KingOfActing

lukewarm mess

InteGrand

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