Exponential and logarithmic functions
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Exponential and logarithmic functions

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Had to make this dumb powerpoint for my algebra II class and I put a lot of work into it for some reason... so yeah it's just been sitting on my laptop doing nothing and I thought why not upload this ...

Had to make this dumb powerpoint for my algebra II class and I put a lot of work into it for some reason... so yeah it's just been sitting on my laptop doing nothing and I thought why not upload this to help other people? So yeah, hope you guys find it useful...

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Exponential and logarithmic functions Exponential and logarithmic functions Presentation Transcript

  • 3-5 – Graphing Exponential Functions6-8 – Graphing Logarithmic Functions9– Solving Exponential Equations10-13– Properties of Logarithms (Product Property, Quotient Property, Power Property) and Solving Logarithmic Equations
  • • ▫ 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.
  • • 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
  • • 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.
  • ▫ 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)
  • • 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
  • 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!
  • • logb x + logb y = logb xyProduct 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/yQuotient 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.
  • Ex: log3 9 + log3 x = log3 451) 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 equationshould look like this : log3 (9x) = log3 453) 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 9xequal to 45.4) Divide both sides by 9 to get x = 5
  • Ex: log6 x – log6 6 = log6 361) 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 363) 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 = 364) To isolate x, multiply both sides by 6 to get x = 216.
  • 3log2 2 + log2 x2 – log2 4 = log2 501) 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^25) From here we can use the equality property to set the arguments equal to each other and eliminate the logs . 2x^2= 506) Divide both sides by 2 to get x^2 = 25. Then, square root both sides to get x = ±5