![\begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( \frac{x}{y}\right)\implies \log_a(x)-\log_a(y) \end{array}~\hfill \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \underset{\stackrel{\uparrow }{\textit{let's use this one}}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bllll%7D%20%5Ctextit%7BLogarithm%20of%20rationals%7D%20%5C%5C%5C%5C%20%5Clog_a%5Cleft%28%20%5Cfrac%7Bx%7D%7By%7D%5Cright%29%5Cimplies%20%5Clog_a%28x%29-%5Clog_a%28y%29%20%5Cend%7Barray%7D~%5Chfill%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Ctextit%7BLogarithm%20Cancellation%20Rules%7D%20%5C%5C%5C%5C%20log_a%20a%5Ex%20%3D%20x%5Cqquad%20%5Cqquad%20%5Cunderset%7B%5Cstackrel%7B%5Cuparrow%20%7D%7B%5Ctextit%7Blet%27s%20use%20this%20one%7D%7D%7D%7Ba%5E%7Blog_a%20x%7D%3Dx%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

notice, -2 is a valid value for the quadratic, however, the argument value for a logarithm can never 0 or less, it has to be always greater than 0, so for the logarithmic expression with (x-2), using x = -2 will give us a negative value, so -2 is no dice.
Answer: 
Step-by-step explanation:
The first four ordered pairs using the two sequences are:
The first rule is to add 10, then divide by 2 starting from 2
Add/Divide:

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Add/divide again:

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Add/divide again:

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Answer:
see explanation
Step-by-step explanation:
Given f(x) then f(x + a) is a horizontal translation of f(x)
• If a > 0 then a shift to the left of a units
• If a < 0 then a shift to the right of a units
Given f(x) then f(x) + c is a vertical translation of f(x)
• If c > 0 then a vertical shift up of c units
• If c < 0 then a vertical shift down of c units
Given
g(x) = (x + 5)² + 2
This represents a shift to the left of 5 units and a shift up of 2 units
Answer:
they are both money
Step-by-step explanation:
that's why they relate, they're both currency
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