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Logarithmic Equations

Fundamental Properties

Fundamental Properties
With a little earthquake research, you could find information on the Richter scale, which was developed by Gütenberg and Richter. The formula relating the energy E (in ergs) to the magnitude of the earthquake M is given by This equation is called a logarithmic equation, and the topic of this section is to solve such equations.

Fundamental Properties
To answer the question, we first solve for log E, and then we use the definition of logarithm to write this as an exponential equation. 1.5M = log E – 11.8 1.5M = log E 101.5M = E Given equation Multiply both sides by 1.5. Add 11.8 to both sides. Definition of logarithm

Fundamental Properties
We can now answer the question. Since M = 7.6, E = 101.5(7.6)  1.58  1023 However, to solve certain logarithmic equations, we must first develop some properties of logarithms.

Fundamental Properties
We begin with two fundamental properties of logarithms. If you understand the definition of logarithm, you can see that these two properties are self-evident, so we call these the Grant’s tomb properties of logarithms.

Example 1 – Evaluate a logarithmic expression
Evaluate the given expressions. a. log 103 b. ln e2 c. 10log 8.3 d. eln 4.5 e. log8 84.2

Example 1 – Solution a. log 103 = 3 b. ln e2 = 2 c. 10log 8.3 = 8.3
d. eln 4.5 = 4.5 e. log = 4.2 log 103 is the exponent on 10 that gives 103; obviously it is 3. ln e2 is the exponent on e that gives e2; obviously it is 2. Consider log 8.3; what is this? It is the exponent on a base 10 that gives the answer 8.3. Consider ln 4.5; what is this? It is the exponent on a base e that gives the answer 4.5. What is the exponent on 8 that gives 84.2? Obviously, it is 4.2.

Logarithmic Equations

Logarithmic Equations
A logarithmic equation is an equation for which there is a logarithm on one or both sides. The key to solving logarithmic equations is the following theorem, which we will call the log of both sides theorem.

Laws of Logarithms

Laws of Logarithms To simplify logarithmic expressions, we remember that a logarithm is an exponent and the laws of exponents correspond to the laws of logarithms.

Laws of Logarithms The proofs of these laws of logarithms are easy. The additive law of logarithms comes from the additive law of exponents: bxby = bx + y Let A = bx and B = by, so that AB = bx + y. Then, from the definition of logarithm, these three equations are equivalent to x = logb A, y = logb B, and x + y = logb(AB)

Laws of Logarithms Therefore, by putting these pieces together, we have logb(AB) = x + y = logb A + logb B Similarly, for the subtractive law of logarithms, Subtractive law of exponents Definition of logarithm

Laws of Logarithms The proof of the multiplicative law of logarithms follows from the multiplicative law of exponents and you are asked to do this in the problem set. We can also prove this multiplicative law by using the additive law of logarithms for p a positive integer: logb Ap = logb (A  A  A      A) = logb A + logb A + logb A +    + logb A Since x = logb A and y = logb B Definition of Ap Additive law of logarithms

Laws of Logarithms When logarithms were used for calculations, the laws of logarithms were used to expand an expression such as log Calculators have made such problems obsolete. Today, logarithms are important in solving equations, and the procedure for solving logarithmic equations requires that we take an algebraic expression involving logarithms and write it as a single logarithm. We might call this contracting a logarithmic expression.

Example 3 – Use logarithmic properties to contract
Write each statement as a single logarithm. a. log x + 5 log y – log z b. log2 3x – 2 log2 x + log2(x + 3) Solution: a. log x + 5 log y – log z = log x + log y 5 – log z = log xy 5 – log z = Third law First law Second law

Example 3 – Solution cont’d b.

Example 6 – Prove change of base theorem
Solution: Let y = loga x. ay = x logb ay = logb x y logb a = logb x Definition of logarithm Log of both sides theorem Third law of logarithms

Example 6 – Solution Thus, by substitution, . cont’d
Divide both sides by logb a (logb a  0).