Define and Identify Exponential and Logarithmic Properties
Learning Outcomes

 Define the identity and zero properties of exponents and logarithms.
 Define the inverse property of exponents and logarithms.
 Define the onetoone properties of exponents and logarithms.
 Define the product and quotient properties of exponents and logarithms.
[latex]\log_b\left(x\right)=y\ \Leftrightarrow \ {b}^{y}=x,\ \text{}b>0,\ b\ne 1[/latex]
That is, to say that the logarithm to base b of [latex]x[/latex] is [latex]y[/latex] is equivalent to saying that [latex]y[/latex] is the exponent on the base b that produces [latex]x[/latex]. Note that the base b is always a positive number other than [latex]1[/latex] and that the logarithmic and exponential forms "undo" each other. This means that logarithms have properties similar to exponents. Let's compare several such properties side by side.Identity and Zero Properties of Exponents and Logarithms
Zero Exponent Rule for Logarithms and Exponentials
Recall that for exponents,[latex]b^0=1[/latex].
Using the definition of a logarithm, we can rewrite that statement in logarithm form:[latex]\log_b1=0[/latex]
Identity Exponent Rule for Logarithms and Exponentials
Recall that for exponents,[latex]b^1 = b[/latex]
Using the definition of a logarithm, we can rewrite that statement in logarithm form:[latex]\log_bb=1[/latex]
Example
Use the the fact that exponentials and logarithms are inverses to prove the zero and identity exponent rule for the following:
1. [latex]{\mathrm{log}}_{5}1=0[/latex]
2. [latex]{\mathrm{log}}_{5}5=1[/latex]
Answer:
1.[latex]{\mathrm{log}}_{5}1=0[/latex] since [latex]{5}^{0}=1[/latex]
2.[latex]{\mathrm{log}}_{5}5=1[/latex] since [latex]{5}^{1}=5[/latex]
Inverse Property of Logarithms and Exponents
Exponential and logarithmic functions are inverses of each other, and we can take advantage of this to evaluate and solve expressions and equations involving logarithms and exponentials. The inverse property of logarithms and exponentials gives us an explicit way to rewrite an exponential as a logarithm or a logarithm as an exponential.
Inverse Property of Logarithms and Exponentials
Recall that we can use the definition of a logarithm to rewrite it as an exponent. That is[latex]\log_bx=y \ \Leftrightarrow b^y=x[/latex].
By extension,[latex]\log_bb^x=x \qquad[/latex] and [latex]\qquad b^{\log_bx}=x[/latex].
Example
Evaluate: 1.[latex]\mathrm{log}\left(100\right)[/latex] 2.[latex]{e}^{\mathrm{ln}\left(7\right)}[/latex]Answer: 1. Rewrite the logarithm as [latex]{\mathrm{log}}_{10}\left({10}^{2}\right)[/latex], and then apply the inverse property [latex]{\mathrm{log}}_{b}\left({b}^{x}\right)=x[/latex] to get [latex]{\mathrm{log}}_{10}\left({10}^{2}\right)=2[/latex]. 2. Rewrite the logarithm as [latex]{e}^{{\mathrm{log}}_{e}7}[/latex], and then apply the inverse property [latex]{b}^{{\mathrm{log}}_{b}x}=x[/latex] to get [latex]{e}^{{\mathrm{log}}_{e}7}=7[/latex]
OnetoOne Property of Logarithms and Exponents
Consider the mathematical statement [latex]2^x=2^y[/latex]. The only possible way this could be a true statement is if the variables [latex]x[/latex] and [latex]y[/latex] are equivalent. This statement is an example of the onetoone property of exponents. More formally, it states[latex]a^m = a^n \Rightarrow m=n[/latex].
In more general terms, if like bases raised to perhaps different exponents are given to be equivalent, then the exponents themselves must also be equivalent. The same idea applies to logarithms. If two logarithms of like base but with perhaps different arguments are given to be equivalent, then the arguments themselves must also be equivalent. More formally, the onetoone property of logarithms states[latex]\log_b(M) = \log_b(N) \Rightarrow M = N[/latex].
Example
Solve the equation [latex]{\mathrm{log}}_{3}\left(3x\right)={\mathrm{log}}_{3}\left(2x+5\right)[/latex] for [latex]x[/latex].Answer: In order for this equation to be true, we must find a value for x such that [latex]3x=2x+5[/latex] [latexdisplay]\begin{array}{c}3x=2x+5\hfill & \text{Set the arguments equal to each other}\text{.}\hfill \\ x=5\hfill & \text{Subtract 2}x\text{.}\hfill \end{array}[/latexdisplay] Check your answer by substituting [latex]5[/latex] for [latex]x[/latex].
[latex]\begin{array}{c}{\mathrm{log}}_{3}\left(3\cdot5\right)={\mathrm{log}}_{3}\left(2\cdot5+5\right)\\{\mathrm{log}}_{3}\left(15\right)={\mathrm{log}}_{3}\left(15\right)\end{array}[/latex]
This is a true statement, so we must have found the correct value for [latex]x[/latex].The Product and Quotient Rules for Logarithms
Recall that we use the product rule of exponents to combine the product of like bases raised to exponents by adding the exponents:[latex]{x}^{a}{x}^{b}={x}^{a+b}[/latex].
We have a similar property for logarithms, called the product rule for logarithms, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.The Product Rule for Logarithms
The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.[latex]{\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)[/latex]
Example
Using the product rule for logarithms, rewrite the logarithm of a product as the sum of logarithms of its factors. [latexdisplay]{\mathrm{log}}_{b}\left(wxyz\right)[/latexdisplay]Answer: [latexdisplay]{\mathrm{log}}_{b}\left(wxyz\right)={\mathrm{log}}_{b}w+{\mathrm{log}}_{b}x+{\mathrm{log}}_{b}y+{\mathrm{log}}_{b}z[/latexdisplay]
[latex]\dfrac{x^a}{x^b}={x}^{ab}[/latex].
The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.The Quotient Rule for Logarithms
The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.Example
Expand the following expression using the quotient rule for logarithms. [latexdisplay]\mathrm{log}\left(\dfrac{2{x}^{2}+6x}{3x+9}\right)[/latexdisplay]Answer:
Factoring and canceling, we get:
Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.
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 Precalculus. Provided by: OpenStax Authored by: Jay Abramson, et al.. Located at: https://openstax.org/books/precalculus/pages/1introductiontofunctions. License: CC BY: Attribution. License terms: Download For Free at : http://cnx.org/contents/[email protected]..