{"id":1195,"date":"2024-04-15T14:43:22","date_gmt":"2024-04-15T14:43:22","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&p=1195"},"modified":"2024-08-05T12:29:58","modified_gmt":"2024-08-05T12:29:58","slug":"exponential-and-logarithmic-functions-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/exponential-and-logarithmic-functions-learn-it-2\/","title":{"raw":"Exponential and Logarithmic Functions: Learn It 2","rendered":"Exponential and Logarithmic Functions: Learn It 2"},"content":{"raw":"

Exponential Functions Cont.<\/h2>\r\n

Evaluating Exponential Functions<\/h3>\r\n

To evaluate an exponential function with the form [latex]f(x)=b^x[\/latex], we simply substitute [latex]x[\/latex] with the given value, and calculate the resulting power.<\/p>\r\n

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Let [latex]f(x)=2^x[\/latex]. What is [latex]f(3)[\/latex]?<\/p>\r\n

[latex]\\begin{array}{rcl} f(x) & = & 2^x \\\\ f(3) & = & 2^3 & \\quad \\text{Substitute } x = 3. \\\\ & = & 8 & \\quad \\text{Evaluate the power.} \\end{array} [\/latex]<\/center>\r\n

 <\/p>\r\n<\/section>\r\n

To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations.<\/p>\r\n

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Let [latex]f(x)=30(2)^x[\/latex]. What is [latex]f(3)[\/latex]?<\/p>\r\n

[latex]\\begin{array}{rcll} f(x) & = & 30(2)^x & \\\\ f(3) & = & 30(2)^3 & \\quad \\text{Substitute } x = 3. \\\\ & = & 30(8) & \\quad \\text{Simplify the power first.} \\\\ & = & 240 & \\quad \\text{Multiply.} \\end{array} [\/latex]<\/center>Note that if the order of operations were not followed, the result would be incorrect:
[latex]f(3)=30(2)^3\u226060^3=216,000[\/latex]<\/center><\/section>\r\n
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How To: Evaluating Exponential Functions<\/b><\/p>\r\n

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  1. Given an exponential function, identify [latex]a[\/latex], [latex]b[\/latex], and the value of [latex]x[\/latex] you're being asked to substitute into the function.<\/li>\r\n\t
  2. Replace the variable [latex]x[\/latex] in the function with the given number.<\/li>\r\n\t
  3. Compute the value of [latex]b^x[\/latex]. This means raising the base [latex]b[\/latex] to the power of [latex]x[\/latex].<\/li>\r\n\t
  4. If there is a coefficient [latex]a[\/latex] in front of the base, multiply the result of [latex]b^x[\/latex] by [latex]a[\/latex]. If [latex]a[\/latex] is [latex]1[\/latex], this step does not change the value.<\/li>\r\n\t
  5. Simplify the expression if necessary. This could involve performing any additional multiplication or addition\/subtraction if the function has more terms.<\/li>\r\n<\/ol>\r\n<\/section>\r\n
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    Let [latex]f(x)=5(3)^x+1[\/latex]. Evaluate [latex]f(2)[\/latex] without using a calculator.<\/p>\r\n


    \r\n[reveal-answer q=\"586760\"]Show Answer[\/reveal-answer]
    \r\n[hidden-answer a=\"586760\"]Follow the order of operations. Be sure to pay attention to the parentheses.<\/p>\r\n

    [latex]\\begin{array}{rcll} f(x) & = & 5(3)^{x+1} & \\\\ f(2) & = & 5(3)^{2+1} & \\quad \\text{Substitute } x = 2. \\\\ & = & 5(3)^3 & \\quad \\text{Add the exponents.} \\\\ & = & 5(27) & \\quad \\text{Simplify the power.} \\\\ & = & 135 & \\quad \\text{Multiply.} \\end{array} [\/latex]<\/center>\r\n

    [\/hidden-answer]<\/p>\r\n<\/section>\r\n

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    [ohm_question hide_question_numbers=1]284250[\/ohm_question]<\/p>\r\n<\/section>\r\n

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    Suppose a particular population of bacteria is known to double in size every [latex]4[\/latex] hours. If a culture starts with [latex]1000[\/latex] bacteria, the number of bacteria after [latex]4[\/latex] hours is [latex]n(4)=1000\u00b72[\/latex]. The number of bacteria after [latex]8[\/latex] hours is [latex]n(8)=n(4)\u00b72=1000\u00b72^2[\/latex].<\/p>\r\n

    In general, the number of bacteria after [latex]4m[\/latex] hours is [latex]n(4m)=1000\u00b72^m[\/latex]. Letting [latex]t=4m[\/latex], we see that the number of bacteria after [latex]t[\/latex] hours is [latex]n(t)=1000\u00b72^{t\/4}[\/latex].<\/p>\r\n

    Find the number of bacteria after [latex]6[\/latex] hours, [latex]10[\/latex] hours, and [latex]24[\/latex] hours.<\/p>\r\n

    [reveal-answer q=\"fs-id1170572550969\"]Show Solution[\/reveal-answer]
    \r\n[hidden-answer a=\"fs-id1170572550969\"]<\/p>\r\n

    The number of bacteria after [latex]6[\/latex] hours is given by [latex]n(6)=1000\u00b72^{6\/4} \\approx 2828[\/latex] bacteria.<\/p>\r\n

    The number of bacteria after [latex]10[\/latex] hours is given by [latex]n(10)=1000\u00b72^{10\/4} \\approx 5657[\/latex] bacteria.<\/p>\r\n

    The number of bacteria after [latex]24[\/latex] hours is given by [latex]n(24)=1000\u00b72^{24\/4}=1000\u00b72^6=64,000[\/latex] bacteria.<\/p>\r\n

    [\/hidden-answer]<\/p>\r\n<\/section>\r\n

    Laws of Exponents<\/h3>\r\n

    The Laws of Exponents are fundamental rules that govern the operations involving powers. These rules are essential for simplifying expressions and are foundational for higher-level math.<\/p>\r\n

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    laws of exponents<\/h3>\r\n
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    1. The Product of Powers<\/strong> rule states that when you multiply two exponents with the same base, you can add the exponents.
      [latex]b^x\u00b7b^y=b^{x+y}[\/latex]<\/center><\/li>\r\n\t
    2. The Quotient of Powers<\/strong> rule tells us that when dividing exponents with the same base, we subtract the exponents.
      [latex]\\large\\frac{b^x}{b^y} \\normalsize = b^{x-y}[\/latex]<\/center><\/li>\r\n\t
    3. The Power of a Power<\/strong> rule shows that when taking an exponent to another exponent, we multiply the exponents.
      [latex](b^x)^y=b^{xy}[\/latex]<\/center><\/li>\r\n\t
    4. The Power of a Product<\/strong> rule lets us know that when raising a product to an exponent, each factor in the product is raised to the exponent.
      [latex](ab)^x=a^x b^x[\/latex]<\/center><\/li>\r\n\t
    5. The Power of a Quotient<\/strong> rule indicates that when a quotient is raised to an exponent, both the numerator and the denominator are raised to the exponent.
      [latex]\\dfrac{a^x}{b^x} =\\left(\\dfrac{a}{b}\\right)^x[\/latex]<\/center><\/li>\r\n<\/ol>\r\n

      Note: This is true for any constants [latex]a>0, \\, b>0[\/latex], and for all [latex]x[\/latex] and [latex]y[\/latex]<\/em><\/p>\r\n<\/section>\r\n

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      Use the laws of exponents to simplify each of the following expressions.<\/p>\r\n

        \r\n\t
      1. [latex]\\large \\frac{(2x^{2\/3})^3}{(4x^{-1\/3})^2}[\/latex]<\/li>\r\n\t
      2. [latex]\\large \\frac{(x^3 y^{-1})^2}{(xy^2)^{-2}}[\/latex]<\/li>\r\n<\/ol>\r\n

        [reveal-answer q=\"fs-id1170572453127\"]Show Solution[\/reveal-answer]
        \r\n[hidden-answer a=\"fs-id1170572453127\"]<\/p>\r\n

          \r\n\t
        1. We can simplify as follows:
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          [latex]\\large \\frac{(2x^{2\/3})^3}{(4x^{-1\/3})^2} \\normalsize = \\large \\frac{2^3(x^{2\/3})^3}{4^2(x^{-1\/3})^2} \\normalsize = \\large \\frac{8x^2}{16x^{-2\/3}} \\normalsize = \\large \\frac{x^2x^{2\/3}}{2} \\normalsize = \\large \\frac{x^{8\/3}}{2}[\/latex]<\/div>\r\n<\/li>\r\n\t
        2. We can simplify as follows:
          \r\n
          [latex]\\large \\frac{(x^3y^{-1})^2}{(xy^2)^{-2}} \\normalsize = \\large \\frac{(x^3)^2(y^{-1})^2}{x^{-2}(y^2)^{-2}} \\normalsize = \\large \\frac{x^6y^{-2}}{x^{-2}y^{-4}} \\normalsize = x^6x^2y^{-2}y^4 = x^8y^2[\/latex]<\/div>\r\n<\/li>\r\n<\/ol>\r\n\r\n\r\nWatch the following video to see the worked solution to this example.