$$y=\log_4\left(x+7\right)-1$$  

$$y=\log_{\left(x-1\right)}4-7$$  

$$y=\log_4\left(x+1\right)-7$$  

$$y=\log_4\left(x-8\right)$$  

$$y=\ln\left(\frac{x-6}{2}\right)$$  

$$y=\ln\left(x\right)-4$$  

$$y=\ln\left(x+2\right)-6$$  

$$y=\ln\left(x-6\right)+2$$  

x=51.59

x=7

x=57.60

x=18.20

A=7877.26

A=1069.14

A=14539.32

A=1060.52

$$f^{-1}\left(x\right)=\left(\frac{x}{8}\right)^3+5$$  

$$f^{-1}\left(x\right)=\left(\frac{x}{8}+5\right)^3$$  

$$f^{-1}\left(x\right)=\frac{x^3}{8}+5$$  

$$f^{-1}\left(x\right)=\frac{x^3+5}{8}$$  

the domain is $$\left(-\infty,\infty\right)$$  

The vertical asymptote is x=-3

$$\left(-\infty,\infty\right)$$  The range is

The horizontal asymptote is y=-3

As $$x\rightarrow-\infty$$  ,

$$f\left(x\right)\rightarrow\infty$$  

The domain is $$\left[2,\infty\right)$$   

As $$x\rightarrow\infty$$  , $$f\left(x\right)\rightarrow-1$$  

As $$x\rightarrow-\infty,\ f\left(x\right)\rightarrow\infty$$  

The interval of decrease is $$\left(-\infty,2\right]$$  

The vertical asymptote is x=1

The interval of increase is $$\left(-\infty,\infty\right)$$  

The horizontal asymptote is y=5

As $$x\rightarrow-\infty,\ f\left(x\right)\rightarrow-\infty$$  

x=3

x=8

No solution

x=3,8

x=-2

x=117

x=0

x=7