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This question comes from PURE MATHEMATICS 1 for As and A levels. This question is part of exercise 1 and it has 5 Answers : $1/2, - 2/3, 4, 6 $ and $7$ . The first 2 values $1/2$ and $- 2/3 $ I am able to find but the rest I can't. I get this first 2 values by make 1 to $(x^2 - 11x + 29)^0$ and this solving by transposition as base of power are same so they get cancelled and make it $(6x^2 + x - 2)=0.$

dmtri
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Marz
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5 Answers5

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By inspection, $$a^b=1 \Rightarrow a=1\textrm{ or }b=0 \textrm{ or } a=-1 \textrm{ if } b \textrm{ is even}$$

So we can just solve three different cases.

Case 1: $a=1$. \begin{align}x^2-11x+29&=1\\ x^2-11x+28&=0\\ (x-4)(x-7)&=0\\ x_1&=4\\ x_2&=7\end{align}

Case 2: $b=0$. \begin{align} 6x^2+x-2&=0\\ (3x+2)(2x-1)&=0\\ x_3&=-\frac23\\ x_4&=\frac12 \end{align}

Case 3: $a=-1$ and $b$ is even. \begin{align} x^2-11x+29&=-1\\ x^2-11x+30&=0\\ (x-5)(x-6)&=0\\ x_5&=5 &\textrm{(inadmissible)}\\ x_6&=6 \end{align}

The possible answers are therefore $\boxed{x=-\frac23,\frac12,4,6,7}$.

Andrew Chin
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Don't get tricked.

If $b\ne 0$ and $b \ne \pm 1$ then $b^w = 1$ if and only if $w=0$.

And if $b=1$ then $b^w =1$ no matter what $w$ equals.

And if $b =-1$ then $b^w=1$ if $w$ is even integer or $w$ is a rational number where, in lowest terms, the numerator is even. (we can call that an "even rational".)

And if $b=0$ then $b^w=1$ is impossible. ($0^0$ is undefined.)

So if $b = x^2 -11x + 29$ and if $w = 6x^2+ x-2$ then

$b^w = 1$ if

1) $b= x^2 -11x + 29 =1$

Or if

2) $w= 6x^2+ x-2=0$ and $b=x^2 -11x + 29\ne 0$.

of if

3) $b=x^2-11x + 29 =-1$ and $w= 6x^2+ x-2$ is an "even rational".

So just solve those.

=====

1) $x^2 -11x + 29 =1\implies x^2 -11x + 28 =0 \implies x =\frac {11\pm\sqrt {11^2 -4*28}}2=\frac {11\pm \sqrt{9}}2 = \frac {11-3}2 = 4,7$.

So $x =4$ or $x = 7$ are solutions.

$(4^2 - 11*4+29)^{6*4^2 +4 -2} = 1^{98} = 1$ and $(7^2 -11*7 + 29)^{6*7^2+ 7 -2} = 1^{299} = 1$.

2) $6x^2 +x -2 =0 \implies x=\frac {-1\pm\sqrt{1-4*(-2)*6}}{2*6} = \frac {-1\pm \sqrt {49}}{12} = \frac {-1\pm 7}12 = \frac 12$ or $-\frac 23$

If $x = \frac 12;-\frac 23$ then $x^2 -11x + 29\ne 0$ so these are fine. We have:

$(x^2 -11x + 29)^{6x^2 +x -2}=(x^2-11x+29)^0 = 1$.

3)$x^2 -11x + 29 =-1\implies x^2 -11x + 30 =0\implies x=\frac {11\pm \sqrt{11^2 - 4*30}}{2}=\frac {11\pm 1}2= 5,6$ but this is impossible as $11^2 - 4*30 = -9$.

But we need $6x^2 +x -2$ to be even. $6*25+ 5-2$ is odd, but $6*36+6-2$ is even.

So $x = 6$ is a solution.

$(6^2 -11*6 + 29)^{6*36+6-2} = (-1)^{220} = 1$.

.... so solutions are $x =\frac 12; -\frac 23, 4,7, 6$.

Moo
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fleablood
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Hint: Taking the logarithm on both sides we get $$(6x^2+x-2)\ln(x^2-11x+29)=0$$ so $$6x^2+x-2=0$$ or $$x^2-11x+29=1$$

Dr. Sonnhard Graubner
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1

Hint: few possibilities:

  • $a^{0}=1$ where $a\ne 0$

  • $1^{p}=1$, $\forall p\in \mathbb{R}$

  • $(-1)^{2n}=1$ where $n\in \mathbb{Z}$

Ng Chung Tak
  • 18,990
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It's simple

This equation is true when either $x^2 -11x+29=1$ or $6x^2+x-2 =0$ or $x^2-11x+29=-1 \ \ and \ \ 6x^2+x-2= even$ because $1^k=1$ and $n^0=1$ and $(-1)^{even}=0$

Solve the equations

You have solved second equation so I'll leave that

For the first you will get $$x^2 -11x+28=0$$

$$(x-7)(x-4)=0$$

Solution is$ x=4,7 $as you desired

For third equation

$x^2-11x+30=0\ \ and \ \ 6x^2+x-2= even$

We get

$$x=5,6 \ \ and \ \ 6x^2+x-2=even$$

Plug in each value

For $x=5 6x^2+x-2=153$ which is odd

But for $x=6 6x^2+x-2= 220$ which is even

So x=6 is also a solution

This method is consistent with the As and A Lvl maths

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