Sigma Equation

Same solution with that by Rahul Chopra. Just trying to present it better.

$\log _{343} {7} \sum _{m=1} ^{x!} {m!} = \log _{343} {343^{\frac{1}{3}}} \sum _{m=1} ^{x!} {m!} = \dfrac {1}{3} \sum _{m=1} ^{x!} {m!} = 291 \quad \Rightarrow \sum _{m=1} ^{x!} {m!} = 873$

And since $1!+2!+ 3!+4!+5!+6! = 1+2+6+24+120+720 = 873$ , $\quad \Rightarrow x! = 6\quad \Rightarrow x = \boxed {3}$ .

We know that the summation from m=1 to x! of m! is equal to 291/(log base 343 of 7) = 291/(1/3) = 291*3 = 873. We then need to find the summation of factorials that leads to that number. 1!+2!+3!+4!+5!+6! = 873. Thus x! = 6, x = 3.