1) The document discusses graphing and properties of exponential and logarithmic functions, including: graphing exponential functions by substituting values of the variable into the equation, graphing logarithmic functions using the change of base formula, and properties like the product, quotient, and power properties of logarithms. 2) Examples are provided of solving exponential and logarithmic equations using properties like changing bases to the same value, multiplying or dividing arguments using the product and quotient properties, and applying exponents using the power property. 3) Steps shown include using properties to isolate the variable, set arguments or exponents equal to each other, and solve the resulting equation.

2. 3-5 – Graphing Exponential Functions
6-8 – Graphing Logarithmic Functions
9– Solving Exponential Equations
10-13– Properties of Logarithms (Product
Property, Quotient Property, Power Property)
and Solving Logarithmic Equations

3. •
▫ Its base is a positive real number – except for 1.
▫ It contains a variable in the exponent.
▫ Its domain is all real numbers, and its range is y > 0.
▫ The asymptote for exponential functions will generally be the
x-axis.
▫ Exponential Functions typically resemble this model:
f(x) = b^x
 Ex: f(x) = 6^x
 To graph this particular equation, substitute values for
the exponent (x) and plot the points.

5. • To shift the graph left or right, • To shift the graph up or down,
add or subtract a number from add or subtract a number from
the exponent, respectively. the function, respectively.
• Example: the red curve has • Example: the red curve has
been shifted 5 to the left from been shifted 5 up from the
the graph 6^x ; its function is graph of 6^x ; its function is
6^(x+5) 6^x +5

6. • What is a logarithmic function…?
▫ A logarithmic function is the inverse of an exponential function.
 If an exponential function were reflected across the line y=x, the
reflection would be a logarithmic function.
▫ Logarithms typically resemble this model: x = logb y where x is the
power, b is the base, and y is known as the argument.
▫ Common logarithms have a base of 10 and are written without the base
in the equation: x = log y
▫ Natural logarithms have a base of “e,” a constant which is equal to
2.7182818…; natural logs are noted as ln.
▫ Since logarithms are the inverse of exponential functions, their
properties are also the inverse of exponentials; their domain is x>0, they
have a range of all real numbers, and they approach the y-axis as their
asymptote.

7. ▫ To graph logarithms, you would need to use the “change of base
formula.”
 The change of base formula looks like this: f(x) = logy/logx. To
graph logarithms with technology, using the change of base
formula is imperative as most devices will only graph common
logs (logs with a base of 10). Basically, you would divide the log
of the argument by the log of the base.
 For example, let’s graph f(x) =log12 x. To graph this function, divide
the log of the argument (x) by the log of the base (12) : lg(x)/lg(12)

8. • To shift the graph left or right, • To shift the graph up or down,
add or subtract a number from add or subtract a number from
the argument, respectively. the function, respectively.
• Example: the red curve has • Example: the red curve has
been shifted 6 to the left of been shifted 6 up from log(x);
log(x); its function is log(x+6) its function is log(x) +6

9. 1. Create equal bases for both sides of the equation. Since (√36) is equal to 6^1/2
you can replace it with that; so far your equation should look like this (6^(1/2*(4x-
6)) = 216^2x. Before we do any simplifying on the left side, let’s change the base of
the right side. 216 is the same as 6^3, so replace it with that to get (6^(1/2*(4x-6))
= 6^(3*2x).
2. Now that we have equal bases, simplify the exponents. ½ multiplied by (4x-6) is
(2x-3) so we would have 6^ (2x-3) . On the right side, to simplify, multiply 3 and
2x to get 6^(6x).
3. After we’ve simplified, just eliminate the bases and set the exponent values equal
to each other to solve for the variable. 2x-3 =6x (subtract 2x from both sides).
-3= 3x (divide by 3 to isolate x) x = -1!

10. • logb x + logb y = logb xy
Product Property • When two or more logarithms with the same base are
added their arguments (x and y in this case) are
multiplied.
• logb x - logb y = logb x/y
Quotient Property • When two or more logarithms with the same base are
subtracted their arguments are divided (whichever
comes first in the equation will be the numerator)
• logb x^y = y logb x
Power Property • The exponent on the argument of a logarithm can also
be the coefficient on the logarithmic term.

11. Ex: log3 9 + log3 x = log3 45
1) To solve this equation we must use the power property of logarithms, that means
that the arguments of log3 9 + log3 x will be multiplied.
2) The arguments 9 and x after being multiplied will equal 9x. So far the equation
should look like this : log3 (9x) = log3 45
3) Since the bases on both sides of the equation are equal and there is an equal sign,
according to equality property, the arguments of both logs must be equal. Set 9x
equal to 45.
4) Divide both sides by 9 to get x = 5

12. Ex: log6 x – log6 6 = log6 36
1) To solve for x in this equation we’ll use the quotient property of logarithms.
According to the quotient property, the arguments of the logarithmic terms can be
divided as their bases are equal.
2) To divide the arguments, put the argument x over 6; so far the problem should look
like this: log6 (x/6) = log6 36
3) Since the bases on both sides of the equal sign are the same, we can eliminate the
logs and set the arguments equal to each other. x/6 = 36
4) To isolate x, multiply both sides by 6 to get x = 216.

13. 3log2 2 + log2 x2 – log2 4 = log2 50
1) For this equation, we’ll be using the power property of logarithms. The property
states that the coefficient of a logarithmic term can also be the exponent of the
argument. That means that the 3 in front of log2 2 can be placed as the exponent of
2. Right now the problem should look like this: log2 2^3 + log2 x2 – log2 4= log2 50.
2) By simplifying 2^3 we know that the argument will be 8.
3) Use the product property to multiply the firs two terms of the equation. log2 8 + log2
x2 will become log2 (8 * x^2). 8 times x^2 is 8x^2. So far the problem looks like
this: log2 8x^2 - log2 4 = log2 50.
4) Next use the quotient property to divide the arguments of the remaining two terms.
log2 8x^2 /4 = log2 10. 8x^2 /4 is 2x^2
5) From here we can use the equality property to set the arguments equal to each other
and eliminate the logs . 2x^2= 50
6) Divide both sides by 2 to get x^2 = 25. Then, square root both sides to get x = ±5