We can say that: y = loga(x) is the same as: x = ay. So for all intents and purposes, a logarithm is an exponent. When we see loga (x), we are asking for the exponent to which the base (a) must be raised to obtain (x). As an example, suppose we see: log2 (8). We are asking what exponent must the base (2) be raised to, in order to obtain 8. The answer is 3, since 23 = 8 : log2 (8) = 3. We need to know how to convert between exponential and logarithmic forms. The process is fairly simple, we just need to understand what is being isolated in each scenario. In exponential form: 32 = 9, here 9, the power is isolated. In logarithmic form, we have: log3 (9) = 2, here 2, the exponent is isolated. We solve simple logarithmic equations by converting into exponential form and solving the resulting equation.

Test Objectives
• Demonstrate an understanding of logarithms
• Demonstrate the ability to change between logarithmic form and exponential form
• Demonstrate the ability to solve simple logarithmic equations
• Demonstrate the ability to graph logarithmic functions
• Demonstrate an understanding of function transformations
Logarithmic Functions Practice Test:

#1:

Instructions: evaluate each.

$$a)\hspace{.2em}\log_{343}\left(\frac{1}{7}\right)$$

$$b)\hspace{.2em}\log_{3}\left(\frac{1}{81}\right)$$

$$c)\hspace{.2em}\log_{5}\left(\frac{1}{125}\right)$$

$$d)\hspace{.2em}\log_{4}(64)$$

#2:

Instructions: solve each equation.

$$a)\hspace{.2em}\log_{x}(6)=\frac{1}{3}$$

$$b)\hspace{.2em}\log_{4}(x)=2$$

$$c)\hspace{.2em}\log_{3}(243)=x$$

$$d)\hspace{.2em}\log_{6}(-36)=x$$

#3:

Instructions: solve each equation.

$$a)\hspace{.2em}5 - 9\log_{5}(-4x - 1)=14$$

$$b)\hspace{.2em}4\log_{4}(-3x - 6) - 3=9$$

#4:

Instructions: sketch the graph of f(x) and g(x). Find the transformation from f(x) to g(x).

$$a)\hspace{.2em}f(x)=\log_2(x)$$$$g(x)={-}\log_2(x - 2)$$

$$b)\hspace{.2em}f(x)=\log_{\frac{1}{3}}(x)$$$$g(x)=\log_{\frac{1}{3}}(-x) + 1$$

$$c)\hspace{.2em}f(x)=\log_{2}(x)$$$$g(x)=2\log_{2}(x + 1) - 2$$

#5:

Instructions: find the domain.

$$a)\hspace{.2em}f(x)=\log_5(x^2 + x - 20)$$

$$b)\hspace{.2em}f(x)=\log_2(x^3 + 3x^2 - x - 3)$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}{-}\frac{1}{3}$$

$$b)\hspace{.2em}{-}4$$

$$c)\hspace{.2em}{-}3$$

$$d)\hspace{.2em}3$$

#2:

Solutions:

$$a)\hspace{.2em}x=216$$

$$b)\hspace{.2em}x=16$$

$$c)\hspace{.2em}x=5$$

$$d)\hspace{.2em}\text{No Solution}$$

#3:

Solutions:

$$a)\hspace{.2em}x=-\frac{3}{10}$$

$$b)\hspace{.2em}x=-\frac{70}{3}$$

#4:

Solutions:

a) Shifted 2 units right, reflected across the x-axis

b) Reflected across the y-axis, shifted 1 unit up

c) Shifted 1 unit left, vertically stretched by a factor of 2, shifted down by 2 units

#5:

Solutions:

$$a)\hspace{.2em}\{x | x < -5, x > 4\}$$

$$b)\hspace{.2em}\{x |{-}3 < x < -1, x > 1\}$$