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  1. AP Pre Calculus
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Exponential and Logarithmic Equations and Inequalities

Tom Green

Tom Green

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Next Topic - Logarithmic Function Context and Data Modeling

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Study Guide Overview

This study guide covers exponential and logarithmic equations and inequalities. It explains how to use properties of exponents and logarithms, including the product, quotient, and power rules, and the inverse relationship between these functions to solve equations and inequalities. It also emphasizes checking for extraneous solutions. Finally, it covers finding the inverse of exponential and logarithmic functions, including transformations like horizontal and vertical shifts and stretches/compressions.

#2.13 Exponential and Logarithmic Equations and Inequalities

#Approaching Equations and Inequalities of Exponents & Logs

Properties of exponents, properties of logarithms, and the inverse relationship between exponential and logarithmic functions can be used to simplify and solve equations and inequalities involving exponents and logarithms. These properties can be used to change the form of the equation or inequality, making it easier to solve. ๐Ÿ’•

#๐Ÿช„ The Magic Behind Properties!

Some examples of properties of exponents that can be used to simplify equations include the product of powers property, the quotient of powers property, and the power of a power property. For example, if we have the equation 2^x * 2^y = 2^z, we can use the product of powers property to simplify it to 2^(x+y) = 2^z. ๐ŸŽฉ

Similarly, properties of logarithms such as the product rule, quotient rule, and power rule can be used to simplify equations and inequalities involving logarithms. For example, if we have the equation log_2(x) + log_2(y) = log_2(xy), we can use the product rule to simplify it to log_2(x) + log_2(y) = log_2(x) + log_2(y)

The inverse relationship between exponential and logarithmic functions can also be used to solve equations and inequalities. For example, if we have the equation 2^x = 8, we can use the inverse relationship to write it as log_2(8) = x, and then solve for x.

Memory Aid

Key Log Properties

  • Product Rule: log_b(MN) = log_b(M) + log_b(N)
  • Quotient Rule: log_b(M/N) = log_b(M) - log_b(N)
  • Power Rule: log_b(M^p) = p * log_b(M)
  • Change of Base: log_b(M) = log_c(M) / log_c(b)

Think of logs as 'undoing' exponents. If you see a log equation, try to isolate the log term and then rewrite it in exponential form to solve.

#๐Ÿ˜ฌ Extraneous Solutions

When solving exponential and logarithmic equations, it's important to remember to check for extraneous solutions that may have been generated during the solving process but are not actually valid solutions in the context of the problem or the mathematical constraints of the equation. โŒ

Common Mistake

Remember to check for extraneous solutions, especially with logarithms! Logarithms are only defined for positive arguments. Always plug your solutions back into the original equation to make sure they work.

For example, when solving exponential or logarithmic equations, it's important to consider the domain of the exponential or logarithmic function and make sure that the solution is within that domain.

Extraneous Solutions Example

Log(x+1) + log(x-1) = log8 getting solved and checked for extraneous solutions.

Source: Open Algebra

Inverting Exponential and Logarithmic Functions****

#โคด๏ธ Exponential Functions

The function f(x) = ab^(x-h) + k is a combination of additive and multiplicative transformations applied to an exponential function in the general form y = b^x. The additive transformation shifts the function horizontally by a constant value of h and the multiplicative transformation stretches or compresses the function vertically by a constant factor of a.

Exponential Function Transformation

f(x) = 2^x+1 - 3 and y = -3 graphed on coordinate plane.

Source: Lumen Learning

To find the inverse of this function, we must first reverse the additive transformation by subtracting k from both sides of the equation, giving us y - k = ab^(x-h). Next, we must reverse the multiplicative transformation by dividing both sides by a, resulting in (y-k) / a = b^(x-h). ๐Ÿ”„

To undo the exponential function, we take the natural logarithm of both sides, which gives us ln((y-k) / a) = (x-h) ln(b). Solving for x, we get x = h + ln((y-k) / a) / ln(b)

This is the inverse function, which allows us to find the input value x corresponding to a given output value y. ๐Ÿ˜

#๐Ÿชต Logarithmic Functions

The function f(x) = a log_b (x - h) + k is a combination of additive and multiplicative transformations of a logarithmic function. The function is in general form, where "a" is the multiplicative constant, "b" is the base of the logarithm, "h" is a constant value subtracted from the argument of the logarithm, and "k" is a constant value added to the final output of the function.

Logarithmic Function Transformation

Three graphs displayed on a coordinate plane: y=5log(x+2), y=log(x+2), and y=log(x). A vertical line is drawn at the number negative 2 too.

Source: Lumen Learning

To find the inverse of y = f(x), we must first reverse the additive transformation by subtracting k from both sides of the equation, resulting in y - k = a log_b (x - h). Next, we must reverse the multiplicative transformation by dividing both sides by a, resulting in (y - k)/a = log_b (x - h).

To reverse the logarithmic transformation, we take the base "b" to the power of both sides, resulting in b^((y - k) / a) = x - h. Finally, we reverse the subtraction of h by adding h to both sides, resulting in b^((y - k) / a) + h = x. This is the inverse of the original function, f^-1(x) = b^((y-k)/a) + h. ๐Ÿคฉ

Key Concept

Inverting Functions: The Big Idea

  • To invert a function, swap x and y, then solve for y.
  • Remember to reverse the order of operations. Start with addition/subtraction, then multiplication/division, and finally exponents/logs.
  • The inverse of an exponential function is a logarithmic function, and vice-versa.
Exam Tip

Time-Saving Tip: When solving for x in exponential or logarithmic equations, try to isolate the exponential or log term first. This often makes the rest of the process much easier. Also, be ready to use the change of base formula for logarithms if needed.

Practice Question

Multiple Choice

1. Solve for x: 2^(3x-1) = 32
  a) 1
  b) 2
  c) 3
  d) 4
  Answer: b) 2

2. Solve for x: log_3(x+2) + log_3(x) = 1
  a) -3
  b) -1
  c) 1
  d) 3
  Answer: c) 1

3. What is the inverse of f(x) = 3^(x+1) - 2?
  a) f^-1(x) = log_3(x+2) - 1
  b) f^-1(x) = log_3(x-2) + 1
  c) f^-1(x) = log_3(x+2) + 1
  d) f^-1(x) = log_3(x-1) - 2
  Answer: a) f^-1(x) = log_3(x+2) - 1

Free Response

Consider the function f(x) = 2log_3(x-1) + 4. (a) Find the inverse function, f^-1(x). (b) State the domain and range of f(x). (c) State the domain and range of f^-1(x). (d) Evaluate f^-1(10).

Scoring Breakdown

(a) 2 points: 1 point for swapping x and y, 1 point for correctly solving for y (b) 2 points: 1 point for correct domain, 1 point for correct range (c) 2 points: 1 point for correct domain, 1 point for correct range (d) 1 point for correct evaluation

Solution

(a) y=2log3(xโˆ’1)+4y = 2log_3(x-1) + 4y=2log3โ€‹(xโˆ’1)+4 x=2log3(yโˆ’1)+4x = 2log_3(y-1) + 4x=2log3โ€‹(yโˆ’1)+4 xโˆ’4=2log3(yโˆ’1)x - 4 = 2log_3(y-1)xโˆ’4=2log3โ€‹(yโˆ’1) (xโˆ’4)/2=log3(yโˆ’1)(x-4)/2 = log_3(y-1)(xโˆ’4)/2=log3โ€‹(yโˆ’1) 3((xโˆ’4)/2)=yโˆ’13^((x-4)/2) = y-13((xโˆ’4)/2)=yโˆ’1 y=3((xโˆ’4)/2)+1y = 3^((x-4)/2) + 1y=3((xโˆ’4)/2)+1 fโˆ’1(x)=3((xโˆ’4)/2)+1f^-1(x) = 3^((x-4)/2) + 1fโˆ’1(x)=3((xโˆ’4)/2)+1 (b) Domain of f(x): x > 1 or (1, โˆž) Range of f(x): (-โˆž, โˆž) (c) Domain of f^-1(x): (-โˆž, โˆž) Range of f^-1(x): y > 1 or (1, โˆž) (d) fโˆ’1(10)=3((10โˆ’4)/2)+1=3(6/2)+1=33+1=27+1=28f^-1(10) = 3^((10-4)/2) + 1 = 3^(6/2) + 1 = 3^3 + 1 = 27 + 1 = 28fโˆ’1(10)=3((10โˆ’4)/2)+1=3(6/2)+1=33+1=27+1=28

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