What is the value of $x=??$

$\begin{aligned} \prod_{i=1}^x 2018^i & = 2018 \times 2018^2 \times 2018^3 \times \cdots \times 2018^x \\ & = 2018^{1+2+3+\cdots + x} \\ & = 2018^{\frac 12 x(x+1)} \end{aligned}$

Therefore, $\dfrac {x(x+1)}2 = 253$ $\implies x = \left \lfloor \sqrt{2\times 253} \right \rfloor = \boxed {22}$ .

Relevant wiki: Rules of Exponents

First, consider that $a^b\cdot a^c=a^{b+c}$ . Using this, we know that we are simply trying to solve for $x$ , where the first $x$ integers add up to 253. These are known as the triangular numbers, and the $n$ th triangular number can be written as $\frac{n(n+1)}{2}$ . Now, we set that equal to 253: $\frac{x(x+1)}{2}=253$ $x^2+x=506$ $x^2+x-506=0$ $(x+23)(x-22)=0$ $x=-23;~22$ Since $x$ can't be negative, our answer is 22. $\beta_{\lceil \mid \rceil}$

Relevant wiki: Rules of Exponents

Note that the expression can be written as:

$2018^1 \times 2018^2 \times 2018^3 \dots 2018^x = 2018^{253}\implies \frac{x(x+1)}{2}=253\implies x(x+1)=506 \implies x=22 \text { or } x= -23$ .

We take $x=22$ as an answer.