19. How many different real number solutions are there to the following equation?
$$
\left(x^{2}-8 x+15\right)^{\left(x^{5}-5 x^{3}+4 x\right)}=1
$$
SolUTION
008

Some solutions occur when $x^{5}-5 x^{3}+4 x=0$, provided $x^{2}-8 x+15 \neq 0$ simultaneously. Therefore we consider $x\left(x^{4}-5 x^{2}+4\right)=0$, i.e. $x\left(x^{2}-1\right)\left(x^{2}-4\right)=0$ i.e. $x(x-1)(x+1)(x-2)(x+2)=0$. The solutions to this are $-2,-1,0,1$ and 2 , none of which makes $x^{2}-8 x+15=0$, so all five of these are solutions.
Some solutions occur when $x^{2}-8 x+15=1$. This has two distinct solutions $x=4 \pm \sqrt{2}$. Some solutions may occur when $x^{2}-8 x+15=-1$. This is equivalent to $x^{2}-8 x+16=0$ which has exactly one solution, namely $x=4$. This will only solve the original equation if the index $x^{5}-5 x^{3}+4 x$ is even when $x=4$. At $x=4$, the index has a value of $4^{5}-5 \times 4^{3}+4 \times 4=720$, which is even. Therefore $x=4$ is a solution.
Therefore, there are $5+2+1=8$ different solutions.