How to solve $(x^2 - 11x + 29)^{(6x^2 + x - 2)}=1$? 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 $(6x^2 + x - 2)=0.$ 
 A: 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 $\boxed{x=-\frac23,\frac12,4,6,7}$.
A: 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 $$x^2-11x+29=1$$
A: Hint: few possibilities:


*

*$a^{0}=1$ where $a\ne 0$

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

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