### Instruction: The sequence \[\log_{12}{162}, \ \log_{12}{x}, \ \log_{12}{y}, \ \log_{12}{z}, \ \log_{12}{1250}\]is an arithmetic progression. What is $x$?### Response: To solve this problem, I need to remember what an arithmetic progression is.
It is a sequence of numbers where each term is obtained by adding a constant difference to the previous term.
For example, 3, 5, 7, 9, 11 is an arithmetic progression with a difference of 2.
If I let the difference be $d$, then I can write a general formula for the terms of the arithmetic progression.
The first term is \[\log_{12}{162},\]the second term is \[\log_{12}{162} + d,\]the third term is \[\log_{12}{162} + 2d,\]and so on.
In particular, the fifth term is \[\log_{12}{162} + 4d.\]
I also know that the fifth term is equal to \[\log_{12}{1250}.\]
This gives me an equation that I can use to find $d$:

\[\log_{12}{162} + 4d = \log_{12}{1250}.\]
To simplify this equation, I can use the properties of logarithms.
Specifically, I can use the fact that \[\log_{b}{a} - \log_{b}{c} = \log_{b}{\frac{a}{c}}.\]
This allows me to write

\[4d = \log_{12}{\frac{1250}{162}}.\]
To isolate $d$, I can divide both sides by 4:

\[d = \frac{1}{4} \log_{12}{\frac{1250}{162}}.\]
Now that I have $d$, I can use it to find $x$.
I know that the second term of the arithmetic progression is \[\log_{12}{x},\]and it is equal to

\[\log_{12}{162} + d.\]
Substituting the value of $d$, I get

\[\log_{12}{x} = \log_{12}{162} + \frac{1}{4} \log_{12}{\frac{1250}{162}}.\]
To solve for $x$, I can use the properties of logarithms again.
This time, I can use the fact that \[\log_{b}{a} + \log_{b}{c} = \log_{b}{ac}.\]
This allows me to write

\[\log_{12}{x} = \log_{12}{162 \cdot \left(\frac{1250}{162}\right)^{\frac{1}{4}}}.\]
To get rid of the logarithm, I can exponentiate both sides with base 12:

\[x = 12^{\log_{12}{162 \cdot \left(\frac{1250}{162}\right)^{\frac{1}{4}}}}.\]
Using the property that \[b^{\log_{b}{a}} = a,\]I can simplify the right-hand side to

\[x = 162 \cdot \left(\frac{1250}{162}\right)^{\frac{1}{4}}.\]