Exponential and Logarithmic Equations and Inequalities

#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. 💕

#🪄 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. 🎩

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.

#😬 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. ❌

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.

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

Source: Open Algebra

Inverting Exponential and Logarithmic Functions****

#⤴️ 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.

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). 🔄

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. 😁

#🪵 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.

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. 🤩

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