Solve $a={x}^{n}\text{},\text{}b=(x+1{)}^{n}$ for $x,n$

Garrett Black
2022-06-28
Answered

Solve $a={x}^{n}\text{},\text{}b=(x+1{)}^{n}$ for $x,n$

You can still ask an expert for help

candelo6a

Answered 2022-06-29
Author has **24** answers

$\frac{b}{a}={\left(\frac{x+1}{x}\right)}^{n}$

${\left(\frac{b}{a}\right)}^{\frac{1}{n}}=1+\frac{1}{x}$

$x=\frac{1}{{\left(\frac{b}{a}\right)}^{\frac{1}{n}}-1}$

Substitute this in $(1)$ to find $n$.

Now use since $x$ is an integer and given $x=\frac{1}{{\left(\frac{b}{a}\right)}^{\frac{1}{n}}-1}$, what can you infer?

EDIT: As x is a positive integer as is said by OP in the comments, hence ${\left(\frac{b}{a}\right)}^{\frac{1}{n}}-1=1$, or $b=a{2}^{n}$ or $n=\frac{log\frac{b}{a}}{log2}$.

$\therefore n=\frac{log\frac{b}{a}}{log2}$ and $x=1$

${\left(\frac{b}{a}\right)}^{\frac{1}{n}}=1+\frac{1}{x}$

$x=\frac{1}{{\left(\frac{b}{a}\right)}^{\frac{1}{n}}-1}$

Substitute this in $(1)$ to find $n$.

Now use since $x$ is an integer and given $x=\frac{1}{{\left(\frac{b}{a}\right)}^{\frac{1}{n}}-1}$, what can you infer?

EDIT: As x is a positive integer as is said by OP in the comments, hence ${\left(\frac{b}{a}\right)}^{\frac{1}{n}}-1=1$, or $b=a{2}^{n}$ or $n=\frac{log\frac{b}{a}}{log2}$.

$\therefore n=\frac{log\frac{b}{a}}{log2}$ and $x=1$

asked 2022-07-10

How to solve the following pair of equation

1. ${x}^{2}+12x+{y}^{2}-4y=24$

2. ${x}^{2}-6x+{y}^{2}+8y=25$

1. ${x}^{2}+12x+{y}^{2}-4y=24$

2. ${x}^{2}-6x+{y}^{2}+8y=25$

asked 2022-07-04

How to frame this set of linear equations?

$2x+1y+2z=A$

$0x+2y+2z=A$

$1x+2y+1z=A$

I assume this can be rewritten as a matrix? How can I check if a solution exists such that x, y, and z are nonnegative?

$2x+1y+2z=A$

$0x+2y+2z=A$

$1x+2y+1z=A$

I assume this can be rewritten as a matrix? How can I check if a solution exists such that x, y, and z are nonnegative?

asked 2022-05-16

wo identitical pendula each of length $l$ and with bobs of mass $m$ are free to oscillate in the same plane. The bobs are joined by a spring with spring constant $k$, by looking for solutions where $x$ and $y$ vary harmonically at the same angular frequency $\omega $, form a simultaneous equation for the amplitudes of oscillation ${x}_{0}$ and ${y}_{0}$.

Considering the forces acting on each pendulum we can derive the following coupled-differential equations:

$\begin{array}{}\text{(1)}& m\ddot{x}& =k(y-x)-\frac{mgx}{\ell}\text{(2)}& m\ddot{y}& =-k(y-x)-\frac{mgy}{\ell}\end{array}$

Where $x$ and $y$ are the displacements of each of the pendulum as functions of time. If we assume they oscillate harmonically with angular frequency $\omega $ then we can write $\mathrm{\exists}\omega ,{\varphi}_{1},{\varphi}_{2}\in \mathbb{R}$:

$\begin{array}{rl}x(t)& ={x}_{0}\mathrm{cos}(\omega t+{\varphi}_{1})\\ y(t)& ={y}_{0}\mathrm{cos}(\omega t+{\varphi}_{2})\end{array}$

Substituting these solutions back into $(1)$ and $(2)$ we get:

$\begin{array}{rl}-m{\omega}^{2}{x}_{0}\mathrm{cos}(\omega t+{\varphi}_{1})& =k({x}_{0}\mathrm{cos}(\omega t+{\varphi}_{1})-{y}_{0}\mathrm{cos}(\omega t+{\varphi}_{2}))-\frac{mg}{\ell}\mathrm{cos}(\omega t+{\varphi}_{1})\\ -m{\omega}^{2}{y}_{0}\mathrm{cos}(\omega t+{\varphi}_{2})& =-k({x}_{0}\mathrm{cos}(\omega t+{\varphi}_{1})-{y}_{0}\mathrm{cos}(\omega t+{\varphi}_{2}))-\frac{mg}{\ell}\mathrm{cos}(\omega t+{\varphi}_{2})\end{array}$

However, without assuming that ${\varphi}_{1}={\varphi}_{2}$, in which case everything factors out nicely to leave a simultaneous equation in ${x}_{0}$ and ${y}_{0}$, I cannot see a way of making it linear in ${x}_{0}$ and ${y}_{0}$. So am I expected to use this assumption or is there a mathematical way of simplifying it?

If it is the former, then what would the physical justification for this assumption be?

Considering the forces acting on each pendulum we can derive the following coupled-differential equations:

$\begin{array}{}\text{(1)}& m\ddot{x}& =k(y-x)-\frac{mgx}{\ell}\text{(2)}& m\ddot{y}& =-k(y-x)-\frac{mgy}{\ell}\end{array}$

Where $x$ and $y$ are the displacements of each of the pendulum as functions of time. If we assume they oscillate harmonically with angular frequency $\omega $ then we can write $\mathrm{\exists}\omega ,{\varphi}_{1},{\varphi}_{2}\in \mathbb{R}$:

$\begin{array}{rl}x(t)& ={x}_{0}\mathrm{cos}(\omega t+{\varphi}_{1})\\ y(t)& ={y}_{0}\mathrm{cos}(\omega t+{\varphi}_{2})\end{array}$

Substituting these solutions back into $(1)$ and $(2)$ we get:

$\begin{array}{rl}-m{\omega}^{2}{x}_{0}\mathrm{cos}(\omega t+{\varphi}_{1})& =k({x}_{0}\mathrm{cos}(\omega t+{\varphi}_{1})-{y}_{0}\mathrm{cos}(\omega t+{\varphi}_{2}))-\frac{mg}{\ell}\mathrm{cos}(\omega t+{\varphi}_{1})\\ -m{\omega}^{2}{y}_{0}\mathrm{cos}(\omega t+{\varphi}_{2})& =-k({x}_{0}\mathrm{cos}(\omega t+{\varphi}_{1})-{y}_{0}\mathrm{cos}(\omega t+{\varphi}_{2}))-\frac{mg}{\ell}\mathrm{cos}(\omega t+{\varphi}_{2})\end{array}$

However, without assuming that ${\varphi}_{1}={\varphi}_{2}$, in which case everything factors out nicely to leave a simultaneous equation in ${x}_{0}$ and ${y}_{0}$, I cannot see a way of making it linear in ${x}_{0}$ and ${y}_{0}$. So am I expected to use this assumption or is there a mathematical way of simplifying it?

If it is the former, then what would the physical justification for this assumption be?

asked 2022-07-09

Is it wrong to write a linear system as below?

Suppose we have the following linear system

$\begin{array}{rl}2{x}_{1}+3{x}_{2}+4{x}_{3}& =1\\ 2{x}_{2}-3{x}_{3}& =6\\ 0& =0.\end{array}$

Is it wrong to write the above linear system by including all the zero coefficients as below

$\begin{array}{rl}2{x}_{1}+3{x}_{2}+4{x}_{3}& =1\\ 0{x}_{1}+2{x}_{2}-3{x}_{3}& =6\\ 0{x}_{1}+0{x}_{2}+0{x}_{3}& =0\end{array}$

instead?

Suppose we have the following linear system

$\begin{array}{rl}2{x}_{1}+3{x}_{2}+4{x}_{3}& =1\\ 2{x}_{2}-3{x}_{3}& =6\\ 0& =0.\end{array}$

Is it wrong to write the above linear system by including all the zero coefficients as below

$\begin{array}{rl}2{x}_{1}+3{x}_{2}+4{x}_{3}& =1\\ 0{x}_{1}+2{x}_{2}-3{x}_{3}& =6\\ 0{x}_{1}+0{x}_{2}+0{x}_{3}& =0\end{array}$

instead?

asked 2022-09-12

Solve the system of equations in response write down the sum of the solutions of the system {4x-2y=2 and below 2x+y=5

asked 2022-05-07

Solve for the variables y and z in the system of equations:

$\{\begin{array}{l}6y+2z=-5\\ 4y+2z=3\end{array}$

$\{\begin{array}{l}6y+2z=-5\\ 4y+2z=3\end{array}$

asked 2021-09-14

Solve the system of equations by hand.