Question

7 multiple choice 1 point what is the asymptote of the graph of ( g(x)=ln x + 5 ) vertical at ( x = e ) vertical at ( x = 0 ) horizontal at ( y = e ) horizontal at ( y = 5 ) 8 matching 1 point complete the equation of the inverse of the function ( f(x)=log (2 x) ) ( f^{-1}(x)=a(b)^{x} ) 9 multiple answer 1 point a new kind of rocket takes off with exponential acceleration with height in miles represented by ( 3^{2} ), where ( x ) is time in minutes. find the time when the rockets height is 8 miles. select the exact solution as a logarithm and select an approximate solution. ( x=log _{2} 3 ) ( x=log _{3} 8 ) ( x=log 8 ) ( x = 1.893 ) ( x = 0.528 ) ( x = 0.903 )

Explanation:

Question 7

Step1: Recall the domain of \(y = \ln x\)

The domain of \(y=\ln x\) is \(x>0\). As \(x\to0^{+}\), \(y = \ln x\to-\infty\).

Step2: Analyze the transformation of \(y=\ln x + 5\)

The function \(g(x)=\ln x+5\) is a vertical shift of \(y = \ln x\) by 5 units up. The vertical asymptote is determined by the domain restriction of the logarithmic function. Since the domain of \(g(x)\) is still \(x > 0\), the vertical asymptote is \(x = 0\).

Step1: Start with \(y=\log(2x)\)

Interchange \(x\) and \(y\): \(x=\log(2y)\)

Step2: Rewrite in exponential form

Using the definition \(y=\log_a u\Leftrightarrow a^y=u\) (here \(a = 10\)), we get \(10^x=2y\)

Step3: Solve for \(y\)

\(y=\frac{10^x}{2}= \frac{1}{2}\times10^x\)

Step1: Set up the equation

We have \(3^x=8\). Using the definition \(a^x = N\Leftrightarrow x=\log_aN\), so \(x=\log_38\)

Step2: Use the change - of - base formula

\(x=\frac{\log8}{\log3}\approx\frac{0.9031}{0.4771}\approx1.893\)

Answer:

vertical at \(x = 0\)

Question 8