Category Archives: Algebra

November 29, 2024 · 8:11 am

Interesting Equation

I think this one is doing the rounds, I first saw it here.

    \begin{equation*}2^x3^{x^2}=6\end{equation}

x=1 is the obvious answer, 2^1\times 3^1=6, but are there more answers?

This was my approach

    \begin{equation*}ln(2^x3^{x^2})=ln(6)\end{equation}

    \begin{equation*}ln(2^x)+ln(3^{x^2})=ln(6)\end{equation}

    \begin{equation*}xln(2)+x^2ln(3)-ln(6)=0\end{equation}

    \begin{equation*}ln(3)x^2+ln(2)x-ln(6)=0\end{equation}

A quadratic equation.

Hence,

    \begin{equation*}x=\frac{-ln(2)\pm\sqrt{(ln(2))^2-4(ln(3))(ln(6))}}{2ln(3)}\end{equation}

I then used my calculator

Hence x=1 0r x=-1.631

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Filed under Algebra, Index Laws, Interesting Mathematics, Quadratics, Solving

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October 7, 2024 · 7:43 am

Arithmetic Sequence

I did this question with on of my year 11 students. I think the algebra and the subscripts can be a bit tricky.

If T_m=n and T_n=m, then prove that T_{m+n}=0. Here where T_n and T_m are terms of an arithmetic sequence.
Mathematics Methods Units 1&2 – Exercise 15B Question 19

If T_m=n then,

(1)   \begin{equation*}n=a+(m-1)d\end{equation*}


And if T_n=m then,

(2)   \begin{equation*}m=a+(n-1)d\end{equation*}


Subtract equation (2) from equation (1)

    \begin{equation*}n-m=(m-1)d-((n-1)d)\end{equation*}


    \begin{equation*}n-m=md-nd\end{equation*}


(3)   \begin{equation*}n-m=d(m-n)\end{equation*}


Therefore d must equal -1
Substitute d=-1 into equation (1)

    \begin{equation*}n=a+(m-1)(-1)\end{equation*}


(4)   \begin{equation*}n=a-m+1\end{equation*}


Therefore a=n+m-1


(5)   \begin{equation*}T_{m+n}=a+(m+n-1)d\end{equation*}


Substitute a=n+m-1 and d=-1 into equation (5)

    \begin{equation*}$T_{m+n}=n+m-1+(m+n-1)(-1)$\end{equation*}


    \begin{equation*}$T_{m+n}=n+m-1-m-n+1$\end{equation*}


(6)   \begin{equation*}$T_{m+n}=0$\end{equation*}

As you can see from equation (6), T_{m+n}=0

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Filed under Algebra, Arithmetic, Sequences, Year 11 Mathematical Methods

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September 23, 2024 · 3:11 am

Cats and Dogs

In my town 10% of the dogs think they are cats and 10% of the cats think they are dogs. All the other cats and dogs are perfectly normal. When all the cats and dogs in my town were rounded up and subjected to a rigorous test, 20% of them thought they were cats. What percentage of them really were cats?
Hamilton Olympiad 2003 B4 – The Ultimate Mathematical Challenge

Let x be the number of cats and y be the number of dogs.
Then 0.9x+0.1y think they are cats.
But we also know 20% of the total think they are cats.
0.2(x+y)
Therefore, 0.9x+0.1y=0.2(x+y)
0.9x+0.1y=0.2x+0.2y
0.7x=0.1y
7x=y
Percentage of cats is \frac{x}{x+y}\times100
Substitute 7x for y
\frac{x}{x+7x}\times100=\frac{x}{8x}\times100=\frac{1}{8}\times100=12.5%
\therefore 12.5% of the animals are cats

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Filed under Algebra, Arithmetic, Percentages, Simplifying fractions, UK Mathematics Challenge

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