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4 multiple answer 1 point the logarithmic function ( g(x)=ln x ) is tra…

Question

4 multiple answer 1 point the logarithmic function ( g(x)=ln x ) is transformed to ( h(x)=ln (x + 2)-1 ). which of the following are true? select all that apply. ( g(x) ) is translated 2 units upward ( g(x) ) is translated 2 units to the right ( g(x) ) is translated 2 units to the left ( g(x) ) is translated 1 unit downward ( g(x) ) is translated 1 unit to the left the vertical asymptote shifts 2 units to the left. the vertical asymptote shifts to units to the right. 5 multiple choice 1 point which of the following represents the inverse of the exponential function ( f(x)=5^{x + 1} ) ( f^{-1}(x)=1+log _{5} x ) ( f^{-1}(x)=log _{5} x-1 ) ( f^{-1}(x)=log _{5}(x - 1) ) ( f^{-1}(x)=log _{5}(x + 1) )

Explanation:

Question 4

Step1: Analyze horizontal translation

For a function \(y = f(x + c)\), it is a horizontal translation. If \(c>0\), the graph of \(y = f(x)\) is translated \(c\) units to the left. For \(h(x)=\ln(x + 2)-1\) compared to \(g(x)=\ln x\), when \(c = 2\) in the \(x\) - part (\(x\to x + 2\)), \(g(x)\) is translated \(2\) units to the left.

Step2: Analyze vertical translation

For a function \(y=f(x)-d\) (\(d>0\)), the graph of \(y = f(x)\) is translated \(d\) units downward. For \(h(x)=\ln(x + 2)-1\) compared to \(g(x)=\ln x\), when \(d = 1\) (since \(h(x)=g(x + 2)-1\)), \(g(x)\) is translated \(1\) unit downward.

Step3: Analyze vertical asymptote

The vertical asymptote of \(y=\ln x\) is \(x = 0\). For \(y=\ln(x + 2)-1\), set \(x+2=0\), we get \(x=-2\). So the vertical asymptote of \(y = g(x)\) (\(x = 0\)) shifts \(2\) units to the left (\(x=-2\))

Step1: Find the inverse of \(y = 5^{x + 1}\)

Start with \(y = 5^{x+1}\). Interchange \(x\) and \(y\): \(x = 5^{y + 1}\)

Step2: Solve for \(y\)

Take the logarithm of both sides with base \(5\). Using the property \(y=\log_{a}a^{z}=z\) (if \(a>0,a
eq1\)), we have \(\log_{5}x=\log_{5}(5^{y + 1})\). By the power rule of logarithms \(\log_{a}M^{n}=n\log_{a}M\), \(\log_{5}x=(y + 1)\log_{5}5\). Since \(\log_{5}5 = 1\), we get \(\log_{5}x=y + 1\)

Step3: Isolate \(y\)

Subtract \(1\) from both sides: \(y=\log_{5}x-1\)

Answer:

  • \(g(x)\) is translated \(2\) units to the left
  • \(g(x)\) is translated \(1\) unit downward
  • The vertical asymptote shifts \(2\) units to the left
Question 5