x

### Warm-up:

### The Problem:

Find $x$.

$${\left(2021\right)}^{0}+{\left(2022\right)}^{1}{\left(2023\right)}^{x}=0$$-/|\-

### Solution:

An easy one to see the new year in with.

The first term, ${\left(2019\right)}^{0}=1$ as any number raised to the zeroth power is one (if it's not zero), and

the second term, ${\left(2020\right)}^{1}=2020$ as any number raised to the power of one is that number.

So, the first two terms sum to

$${\left(2019\right)}^{0}+{\left(2020\right)}^{1}=1+2020=2021$$

Now our equation simplifies to $2021{\left(2021\right)}^{x}=0$,

and further to ${\left(2021\right)}^{x}=2021$

Hence, $x=1$.