Help
Toggle navigation
System Homepage
Advanced Functions and Pre-Calculus
Exponential and Logarithmic Functions
Logarithmic Functions
Watch
Lesson Part 1 (4:43)
Lesson Part 2 (2:06)
Lesson Part 3 (6:30)
Investigation 1
Lesson Part 4 (2:05)
Investigation 2
Lesson Part 5 (5:20)
Alternative Format
Review
01.
02.
03.
04.
05.
06.
07.
Practise
Exercises
Answers
Solutions
Answers
\( \log_5 (125)=3 \)
\( \log_4 (4096)=6 \)
\( \log_{27} \left( 9\sqrt{3}\right)=\frac{5}{6} \)
\( \log_2 \left(\frac{1}{64}\right)=-6 \)
\( 6^3=216 \)
\( 10^3=1000 \)
\( 25^{-2}=\frac{1}{625} \)
\( 16^{\frac{5}{8}}=4\sqrt{2} \)
\( 6 \)
\( 4 \)
\( \frac{1}{2} \)
\( 0 \)
\( -2 \)
\( -3 \)
\( \frac{5}{2} \)
\( \frac{3}{2} \)
\( \frac{5}{3} \)
\( -\frac{3}{2} \)
\( -\frac{3}{2} \)
\( 8 \)
\( \frac{5}{6} \)
\( -4 \)
\( -\frac{3}{2} \)
\( 3.24 \)
\( 7.97 \)
Domain: \( \{x \mid x\in\mathbb{R}\} \), range: \( \{y \mid y \gt 0, y\in\mathbb{R}\} \)
The function \( g(x) = \log_4 (x) \) is the inverse of the function \( f(x)=4^x \). The graph of \( g(x)=\log_4 (x) \) can be obtained by interchanging the \( x \)- and \( y \)-coordinates of the points on \( f(x)=4^x \) and plotting the new points. The graph of \( g(x)=\log_4 (x) \) can also be obtained by reflecting the graph of \( f(x)=4^x \) in the line \( y=x \).
Domain: \( \{x \mid x \gt 0, x\in\mathbb{R}\} \), range: \( \{y \mid y\in\mathbb{R}\} \)
All four functions have a domain \( \{x\in\mathbb{R} \mid x \gt 0\} \), a range \( \{y\in\mathbb{R}\} \), an \( x \)-intercept at \( (1,0) \), and a vertical asymptote of \( x=0 \). \( A(x) \) and \( B(x) \) are decreasing functions and \( C(x) \) and \( D(x) \) are increasing functions.
The graph of \( f(x)=\log_\frac{2}{3} (x) \) will lie above \( B(x)=\log_\frac{1}{2} (x) \) when \( 0\lt x \lt1 \) and below \( B(x)=\log_\frac{1}{2} (x) \) when \( x \gt 1 \), intersecting all four functions at \( (1,0) \). The graph of \( g(x)=\log_5 (x) \) will lie between the graphs of the functions \( C(x)=\log_3 (x) \) and \( D(x)=\log_8 (x) \), intersecting all four functions at \( (1,0) \).
Vertical translation \( 4 \) units down, domain: \( \{x \mid x \gt 0, x\in\mathbb{R}\} \), range: \( \{y \mid y\in\mathbb{R}\} \), vertical asymptote: \( x=0 \)
Reflection in the \( x \)-axis and horizontal translation left \( 3 \) units, domain: \( \{x \mid x \gt -3, x\in\mathbb{R}\} \), range: \( \{y \mid y\in\mathbb{R}\} \), vertical asymptote: \( x=-3 \)
Reflection in the \( y \)-axis and a horizontal stretch about the \( y \)-axis by a factor of \( \frac{1}{2} \), domain: \( \{x \mid x \lt 0, x\in\mathbb{R}\} \), range: \( \{y \mid y\in\mathbb{R}\} \), vertical asymptote: \( x=0 \)
Vertical stretch about the \( x \)-axis by a factor of \( 2 \) and vertical translation \( 3 \) units down, domain: \( \{x \mid x \gt 0, x\in\mathbb{R}\} \), range: \( \{y \mid y\in\mathbb{R}\} \), vertical asymptote: \( x=0 \)
Reflection in the \( x \)-axis, vertical stretch about the \( x \)-axis by a factor of \( 3 \), horizontal translation \( 2 \) units right and vertical translation \( 1 \) unit up, domain: \( \{x \mid x \gt 2, x\in\mathbb{R}\} \), range: \( \{y \mid y\in\mathbb{R}\} \), vertical asymptote: \( x=2 \)
\( g^{-1}(x)=\frac{1}{2} \log_3(x-5)+1 \)
\( g^{-1}(x)=5^{\frac{1}{2}\left(x+1\right)}-4 \)
\( g^{-1}(x)=2\left(3^{4-x}\right)+6 \)
The graphs are identical; therefore, we can conclude that \( \log_2(2x)=\log_2(x)+1 \) for \( x \gt 0, x\in\mathbb{R} \).
The graphs are identical; therefore, we can conclude that \( \log_3\left(\frac{x}{3}\right)=\log_3(x)-1 \) for \( x \gt 0, x\in\mathbb{R} \).
\( (4,2) \)
Previous
More
Quit
Previous Unit Item
Next Unit Item
Next