Unit 9 Seminar Agenda Rational Exponents Logarithmic Functions Properties of Logarithms Exponential and Logarithmic Equations

For review the following are the rules for exponents: 1) xmxn = xm + n product rule 2) (xm)n = xmn power rule 3) (xy)n = xnyn power rule 4) (x/y)n = (xn/yn) power rule 5) x0 = 1; if x ≠ 0 6) (xm/xn) = xm – n quotient rule 7) x-n = (1/xn) negative exponents 8) (x/y)-n = (y/x)n negative exponents

Example 1-using the product rule: Simplify: (x1/2)(x2/3)

Example2-using the power rule: Simplify: (x2)2/3

Example 3: Adding with exponents x1/2 + 5x1/2

Example 4: Multiplying with exponents (add exponents) (3x1/2)(4x1/3)

Example 5: Dividing with exponents

Example 6: Factoring with rational exponents Factor 6x5/2 + 2x3/2 – x1/2.

Example 6: Factoring with rational exponents Factor 6x5/2 + 2x3/2 – x1/2. First we find the GCF The GCF is

Example 6: Factoring with rational exponents Factor 6x5/2 + 2x3/2 –. First we find the GCF The GCF is x1/2. Now we will divide each of the terms by x1/2

= 6x5/2 – ½ = 6x4/2 = 6x2 = 2x3/2 – ½ = 2x2/2 = 2x1 = 2x = 1! Answer: Example 6: Factoring with rational exponents Factor 6x5/2 + 2x3/2 – x1/2. First we find the GCF The GCF is The GCF is x1/2. Now we will divide each of the terms by x1/2 = 6x5/2 – ½ = 6x4/2 = 6x2 = 2x3/2 – ½ = 2x2/2 = 2x1 = 2x = 1! Answer:

= 6x5/2 – ½ = 6x4/2 = 6x2 = 2x3/2 – ½ = 2x2/2 = 2x1 = 2x = 1! Example 6: Factoring with rational exponents Factor 6x5/2 + 2x3/2 – x1/2. First we find the GCF The GCF is The GCF is x1/2. Now we will divide each of the terms by x1/2 = 6x5/2 – ½ = 6x4/2 = 6x2 = 2x3/2 – ½ = 2x2/2 = 2x1 = 2x = 1! Answer: x1/2(6x2 + 2x – 1)

Keep in mind that this can be rewritten like this: Converting between radicals and rational exponents: A rational exponent of 1/n indicated the nth root of the base. In symbolic form it looks like this: For example: So to evaluate a rational exponent, we will change it back to a radical. Example 7: evaluate 163/2 Keep in mind that this can be rewritten like this:

___ ___ 2√163, or just √163 Because radicals and exponents are considered to be the “same level” in the order of operations, you can either deal with the radical first and the exponent second, or the exponent first and the radical second. I will show you both ways.

___ ___ 2√163, or just √163 Because radicals and exponents are considered to be the “same level” in the order of operations, you can either deal with the radical first and the exponent second, or the exponent first and the radical second. I will show you both ways. First evaluating the radical: ___ √163

___ ___ 2√163, or just √163 Because radicals and exponents are considered to be the “same level” in the order of operations, you can either deal with the radical first and the exponent second, or the exponent first and the radical second. I will show you both ways. First evaluating the radical: ___ √163 = 43 = 64 The square root of 16 is 4, and 43 is 4*4*4 = 64 Now we will see what happens when we simplify the exponent first: √163

___ ___ 2√163, or just √163 Because radicals and exponents are considered to be the “same level” in the order of operations, you can either deal with the radical first and the exponent second, or the exponent first and the radical second. I will show you both ways. First evaluating the radical: ___ √163 = 43 = 64 The square root of 16 is 4, and 43 is 4*4*4 = 64 Now we will see what happens when we simplify the exponent first: ___ ____ √163 = √4096 = 64 16 cubed is 16*16*16 = 4096, and the square root of 4096 is 64.

Here are examples of when you might want to convert from radical form to rational exponent form: ___ √x30 = x30/2 = x15 __ 3√y27 = y27/3 = y9

Logarithmic Functions The first thing to remember about a logarithm (log for short) is that a log is an exponent. The definition of a logarithm is as follows: For x > 0 and 0 < a ≠ 1, y = logbx IFF (if and only if) x = by How’s that for textbook gibberish? We will try to decipher what this means....

If we have the equation x = by , how can we solve the equation for the variable y? The answer is:

If we have the equation x = by , how can we solve the equation for the variable y? The answer is: We can’t unless we have some new notation. Hence, the “birth” of the logarithm.

If we have the equation x = by , how can we solve the equation for the variable y? The answer is: We can’t unless we have some new notation. Hence, the “birth” of the logarithm. If we say the equation x = by in words, it will be :

If we have the equation x = by , how can we solve the equation for the variable y? The answer is: We can’t unless we have some new notation. Hence, the “birth” of the logarithm. If we say the equation x = by in words, it will be : ”y is the exponent on the base b needed to get the value x:

If we have the equation x = by , how can we solve the equation for the variable y? The answer is: We can’t unless we have some new notation. Hence, the “birth” of the logarithm. If we say the equation x = by in words, it will be : ”y is the exponent on the base b needed to get the value x: So...if we translate this, remembering that a log is an exponent we can write:

If we have the equation x = by , how can we solve the equation for the variable y? The answer is: We can’t unless we have some new notation. Hence, the “birth” of the logarithm. If we say the equation x = by in words, it will be : ”y is the exponent on the base b needed to get the value x: So...if we translate this, remembering that a log is an exponent we can write: y = logbx

So every exponential equation can be converted to log form and every log equation can be converted to exponential form. Here are the two equivalent equations: Exponential form: x = by Logarithmic form: y = logbx We can use these two equations change the form from one to the other. Let’s do some examples.

Example 8: Convert from log form to exponential form: log381 = 4 is equivalent to log525 = 2 is equivalent to

Example 8: Convert from log form to exponential form: log381 = 4 is equivalent to 34 = 81 log525 = 2 is equivalent to 52 = 25 Example 9: Convert from exponential form to log form: 25 = 32 is equivalent to 42 = 16 is equivalent to

Example 8: Convert from log form to exponential form: log381 = 4 is equivalent to 34 = 81 log525 = 2 is equivalent to 52 = 25 Example 9: Convert from exponential form to log form: 25 = 32 is equivalent to log232 = 5 42 = 16 is equivalent to log416 = 2

Example 10: Now use to solve a log problem where either the b, x or y is missing: log3x = 4 logb16 = 4 log232 = y log48 = y

PROPERTIES OF LOGARITHMS LOG OF A PRODUCT: logaxy = logax + logay Example: log3mn = log3m + log3n Example: log58a = log58 + log5a LOG OF A QUOTIENT: loga(x/y) = logax – logay Example: log8(12/5) = log812 – log85 Example: log6(a/2b) = log6a – log62b LOG OF A POWER: logaxy = ylogax Example: log23y = ylog23 Example: log4x11 = 11log4x

We will try to use the 3 properties to Example: log3w + log3t = log3wt Since the operation was addition, this simplified into the product log. Example: log7a + log7b – log7c = log7(ab/c) Since the operation was addition then subtraction, the first two were multiplied and the third became the divisor (denominator). Example: 4loga3 + 6loga9 = loga34 + loga96 = loga(34*96) (which can later be calculated and simplified further)

Sure, you can probably figure out that the power should be 2, 4x = 16 Sure, you can probably figure out that the power should be 2, but watch the transformation that verifies the equality. 4x = 42 the bases are the same; therefore, x = 2

3x = 1/27 3x = (1/3)3 3x = 3-3 x = -3

this time, set the exponents equal to each other then solve x + 1 = 5

this time, the exponents match, so set the bases equal to each other Example: solve for x: x-3 = 1/1000 x-3 = (1/10)3 x-3 = 10-3 this time, the exponents match, so set the bases equal to each other x = 10

Example: log9x = log95 log9x - log95 = 0 log9(x/5) = 0 (x/5) = 90 (x/5) = 1 5(x/5) = 5(1) x = 5

Example: log15(26) = log15(3x – 1) Since the log’s have the same base You can also just set the inside equal and solve 26 = 3x – 1 +1 + 1 27 = 3x 3 3 9 = x

not have a choice in how to work this one. Note that you do not have a choice in how to work this one. Example: log10x + log105 = 1 log10(x*5) = 1 log10(5x) = 1 5x = 101 5x = 10 5 5 x = 2