### Instruction: Find $x$, given that \[\log_2 2018 + \log_4 2018 + \log_8 2018 + \log_{64} 2018 = \log_x 2018.\].### Response: I notice that the bases of the logarithms are all powers of 2, so I can try to rewrite them using the same base.
Using the change-of-base formula, I have \[\log_2 2018 = \frac{\log_2 2018}{\log_2 2} = \log_2 2018,\] \[\log_4 2018 = \frac{\log_2 2018}{\log_2 4} = \frac{1}{2} \log_2 2018,\] \[\log_8 2018 = \frac{\log_2 2018}{\log_2 8} = \frac{1}{3} \log_.2 2018,\] and \[\log_{64} 2018 = \frac{\log_2 2018}{\log_2 64} = \frac{1}{6} \log_2 2018.\]
So, the left-hand side of the equation becomes \[\log_2 2018 + \frac{1}{2} \log_2 2018 + \frac{1}{3} \log_2 2018 + \frac{1}{6} \log_2 2018 = \frac{6}{6} \log_2 2018 + \frac{3}{6} \log_2 2018 + \frac{2}{6} \log_2 2018 + \frac{1}{6} \log_2 2018 = \frac{12}{6} \log_2 2018 = 2 \log_2 2018.\]
Now, I can compare this to the right-hand side of the equation, which is \[\log_x 2018 = \frac{\log_2 2018}{\log_2 x}.\]
I want to find $x$ such that \[2 \log_2 2018 = \frac{\log_2 2018}{\log_2 x}.\]
I can simplify this by dividing both sides by $\log_2 2018$, assuming it is not zero. I get \[2 = \frac{1}{\log_2 x}.\]
I can solve for $\log_2 x$ by taking the reciprocal of both sides. I get \[\log_2 x = \frac{1}{2}.\]