Logarithmic and exponential functions - Topics in precalculus. Exponential functions Inverse relations Exponential and logarithmic equations Creating one logarithm from a sum THE LOGARITHMIC FUNCTION WITH BASE b is the function y = logb x. b is normally a number greater than 1 (although it need only be greater than 0 and not equal to 1). Note the following: • For any base, the x-intercept is 1. To see the answer, pass your mouse over the colored area. The logarithm of 1 is 0. y = logb1 = 0. • The graph passes through the point (b, 1). The logarithm of the base is 1. logbb = 1. Proper fractions. • The range of the function is all real numbers. • The negative y-axis is a vertical asymptote (Topic 18). Example 1. And here is the graph of y = ln (x − 2) -- which is its translation 2 units to the right.
The x-intercept has moved from 1 to 3. Problem 1. This is a translation 3 units to the left. By definition: logby = x means bx = y. Corresponding to every logarithm function with base b, we see that there is an exponential function with base b: y = bx. Problem 2. and and. Logarithms: Introduction to "The Relationship" Purplemath offers a complete lessonon the topic you have selected.Try the lesson below! This lesson is not yet availablein MathHelp.com. Logarithms: Introduction to "The Relationship" (page 1 of 3) Sections: Introduction to logs, Simplifying log expressions, Common and natural logs Logarithms are the "opposite" of exponentials, just as subtraction is the opposite of addition and division is the opposite of multiplication. Logs "undo" exponentials. In practical terms, I have found it useful to think of logs in terms of The Relationship: On the left-hand side above is the exponential statement "y = bx".
If you can remember this relationship (that whatever had been the argument of the log becomes the "equals" and whatever had been the "equals" becomes the exponent in the exponential, and vice versa), then you shouldn't have too much trouble with logarithms. (I coined the term "The Relationship" myself. Convert "63 = 216" to the equivalent logarithmic expression. Working with Exponents and Logarithms. What is an Exponent? What is a Logarithm? A Logarithm goes the other way. It asks the question "what exponent produced this? ": And answers it like this: In that example: The Exponent takes 2 and 3 and gives 8 (2, used 3 times in a multiplication, makes 8) The Logarithm takes 2 and 8 and gives 3 (2 makes 8 when used 3 times in a multiplication) A Logarithm says how many of one number to multiply to get another number So a logarithm actually gives you the exponent as its answer: (Also see how Exponents, Roots and Logarithms are related.)
Working Together Exponents and Logarithms work well together because they "undo" each other (so long as the base "a" is the same): They are "Inverse Functions" Doing one, then the other, gets you back to where you started: Doing ax then loga gives you x back again: Doing loga then ax gives you x back again: It is too bad they are written so differently ... it makes things look strange. Going up, then down, returns you back again:down(up(x)) = x (and vice versa) And also: CHANGING THE BASE OF A LOGARITHM. Let a, b, and x be positive real numbers such that and (remember x must be greater than 0). Then can be converted to the base b by the formula Let's verify this with a few examples. Example 1: Find to an accuracy of six decimals. , and 7 is between 3 and 9. Can be written When the base is 10, we can leave off the 10 in the notation. . You will note that the answer is between 1 and 2. Let's check the answer. , our answer is correct. .
To six places, so our answer won't check exactly. Example 2: We could work the same problem by converting to the base e. . You will note that the answer is between 1 and 2. Example 3: Find . , and that 18 is between 16 and 32; therefore, we know that the exponent we are looking for is between 4 and 5. Let's solve this problem by changing the base to 10. . To six places, our answer won't check exactly. Example 4: We could work the problem in Example 3 by converting to the base e. The answer is the same as the answer you found when you converted the base to 10. Answer. Basic Log Rules / Expanding Log Expressions. Basic Log Rules / Expanding Logarithmic Expressions (page 1 of 5) Sections: Basic log rules, Expanding, Simplifying, Trick questions, Change-of-Base formula You have learned various rules for manipulating and simplifying expressions with exponents, such as the rule that says that x3 × x5 equals x8 because you can add the exponents.
There are similar rules for logarithms. Log Rules: 1) logb(mn) = logb(m) + logb(n) 2) logb(m/n) = logb(m) – logb(n) 3) logb(mn) = n · logb(m) In less formal terms, the log rules might be expressed as: 1) Multiplication inside the log can be turned into addition outside the log, and vice versa. 2) Division inside the log can be turned into subtraction outside the log, and vice versa. 3) An exponent on everything inside a log can be moved out front as a multiplier, and vice versa. Warning: Just as when you're dealing with exponents, the above rules work only if the bases are the same. Expanding logarithms Expand log3(2x). I have a "2x" inside the log. Log3(2) + log3(x) The Change-of-Base Formula. The Change-of-Base Formula (page 5 of 5) Sections: Basic log rules, Expanding, Simplifying, Trick questions, Change-of-Base formula There is one other log "rule", but it's more of a formula than a rule.
You may have noticed that your calculator only has keys for figuring the values for the common (base-10) log and the natural (base-e) log, but no other bases. Some students try to get around this by "evaluating" something like "log3(6)" with the following keystrokes: Of course, they get the wrong answer, because the above actually calculates the value of "log10(3) × 6". What this rule says, in practical terms, is that you can evaluate a non-standard-base log by converting it to the fraction of the form "(standard-base log of the argument) divided by (same-standard-base log of the non-standard-base)".
This is how you would evaluate the last example on the previous page: Evaluate log3(6). The argument is 6 and the base is 3. Then the answer, rounded to three decimal places, is: log3(6) = 1.631. Logarithms rules and formula. Product rule, power rule, quotient rule and other formulas for logarithms.