1) The first step is to Know the condition for two lines be parallel is that have the same slope, and this form the slope given is 2, therefore the new function has its slope too
2) with the slope and the two points givens, you can find the equation
Y-Y1 = m( X-X1) , Y-6 = 2[(X-(-3)] ⇒Y-6 = 2X+6 ⇒Y = 2X+12, AND THIS IS THE EQUATION PARALLEL TO THE LINE Y = 2X-7
Answer:
4
Step-by-step explanation:
![\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.