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Thread: Solving Exponential Equations

  1. #1
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    Solving Exponential Equations

    Solve:
    $\displaystyle \frac {3}{2}^\frac {m}{2} = \frac {4}{9} $

    How do you cross multiply with this equation?

    ------

    Solve:
    $\displaystyle \frac {27^x}{9^{2x-1}} = 3^{x+4}$

    What do you do here? The first thing I did is putting everything to a common base and distributing the exponents then what?

    Textbook Answer: -1

    -----

    Solve:
    $\displaystyle 10^{z+4} = 10^{z+3} = 11$

    Do you move $\displaystyle 10^{z+3}$ to the right and solve? I did this and I get a different answer from my textbook.

    My answer: $\displaystyle 10^z = \frac {1}{9900}$

    Textbook answer: -3

    -------------------------------

    Solve:
    $\displaystyle 27(3^{3x+1}) = 3 $

    My Work:
    $\displaystyle 3^{3x+1} = \frac {1}{9}$

    $\displaystyle 3^{3x+1} = 3^{-2}$

    $\displaystyle x + 1 = -2$

    $\displaystyle x = -3$

    Textbook Answer: -1

    -------------------------------------------

    Solve:
    $\displaystyle 2^{a+5} + 2^a = 1056$

    My Work:
    $\displaystyle 2^a (2^5 + 2^0) = 1056$

    $\displaystyle 2^a(33) = 1056$

    $\displaystyle 2^a = 2^5$

    $\displaystyle a = 5$

    Textbook Answer: -1

    ----------------------------------------

    Solve:
    $\displaystyle 3^{m+1} + 3^{m+2} - 972 = 0$

    My work:
    $\displaystyle 3^{m+1} (3^0 + 3^{-1}) = 972$

    $\displaystyle 3^{m+1} = 3^6$

    $\displaystyle m = 5$

    Textbook answer: -2

    ----------------------------------------------

    Solve:
    $\displaystyle 5^{n+2} - 5^{n+3} = -2500$

    My Work:
    $\displaystyle 5^n (5^5 - 5^3) = -12500$

    $\displaystyle 5^n = 5^5$

    $\displaystyle n = 2$

    Textbook Answer: 4
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post
    Solve:
    $\displaystyle \frac {3}{2}^\frac {m}{2} = \frac {4}{9} $

    How do you cross multiply with this equation?

    ------
    Is this possibly supposed to be
    $\displaystyle \left ( \frac{3}{2} \right )^{m/2} = \frac{4}{9}$

    Then note that
    $\displaystyle \frac{4}{9} = \left ( \frac{2}{3} \right )^2 = \left ( \frac{3}{2} \right )^{-2}$

    So
    $\displaystyle \left ( \frac{3}{2} \right )^{m/2} = \left ( \frac{3}{2} \right )^{-2}$

    Thus
    $\displaystyle \frac{m}{2} = -2$

    $\displaystyle m = -4$

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post
    Solve:
    $\displaystyle \frac {27^x}{9^{2x-1}} = 3^{x+4}$

    What do you do here? The first thing I did is putting everything to a common base and distributing the exponents then what?

    Textbook Answer: -1

    -----
    You have the right strategy:
    $\displaystyle \frac {27^x}{9^{2x-1}} = 3^{x+4}$

    $\displaystyle \frac {(3^3)^x}{(3^2)^{2x-1}} = 3^{x+4}$

    $\displaystyle \frac {3^{3x}}{3^{2(2x-1)}} = 3^{x+4}$

    $\displaystyle \frac {3^{3x}}{3^{4x-2}} = 3^{x+4}$

    $\displaystyle 3^{3x - (4x - 2)} = 3^{x+4}$

    $\displaystyle 3^{-x + 2} = 3^{x+4}$

    You take it from here.

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post
    Solve:
    $\displaystyle 10^{z+4} = 10^{z+3} = 11$

    Do you move $\displaystyle 10^{z+3}$ to the right and solve? I did this and I get a different answer from my textbook.

    My answer: $\displaystyle 10^z = \frac {1}{9900}$

    Textbook answer: -3
    There's a baaaad typo in here somewhere.

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post
    Solve:
    $\displaystyle 27(3^{3x+1}) = 3 $

    My Work:
    $\displaystyle 3^{3x+1} = \frac {1}{9}$

    $\displaystyle 3^{3x+1} = 3^{-2}$

    $\displaystyle x + 1 = -2$

    $\displaystyle x = -3$

    Textbook Answer: -1
    There's a typo in the third line of your solution.

    Quote Originally Posted by Macleef View Post
    Solve:
    $\displaystyle 2^{a+5} + 2^a = 1056$

    My Work:
    $\displaystyle 2^a (2^5 + 2^0) = 1056$

    $\displaystyle 2^a(33) = 1056$

    $\displaystyle 2^a = 2^5$

    $\displaystyle a = 5$

    Textbook Answer: -1
    I agree with your answer.

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post
    Solve:
    $\displaystyle 3^{m+1} + 3^{m+2} - 972 = 0$

    My work:
    $\displaystyle 3^{m+1} (3^0 + 3^{-1}) = 972$

    $\displaystyle 3^{m+1} = 3^6$

    $\displaystyle m = 5$

    Textbook answer: -2
    I agree with neither answer.
    $\displaystyle 3^{m+1} + 3^{m+2} - 972 = 0$

    $\displaystyle 3^{m+1} + 3^{m+2} = 972$

    $\displaystyle 3^{m + 1} ( 1 + 3^1 ) = 972$ <-- See this line

    $\displaystyle 3^{m + 1}(4) = 972$

    $\displaystyle 3^{m + 1} = 243 = 3^5$

    etc.

    -Dan
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post

    Solve
    :

    $\displaystyle 5^{n+2} - 5^{n+3} = -2500$

    My Work:
    $\displaystyle 5^n (5^5 - 5^3) = -12500$

    $\displaystyle 5^n = 5^5$

    $\displaystyle n = 2$

    Textbook Answer: 4
    Again, I'm not agreeing with the textbook.

    A couple of problems here:
    $\displaystyle 5^{n+2} - 5^{n+3} = -2500$

    $\displaystyle 5^{n+2}(1 - 5^1 ) = -2500$

    $\displaystyle 5^{n + 2}(-4) = -2500$

    $\displaystyle 5^{n + 2} = 625 = 5^4$

    etc.

    I agree with your answer but you had
    $\displaystyle 5^n (5^5 - 5^3) = -12500$
    It's fine to merely pull the $\displaystyle 5^n$ out, but you should have
    $\displaystyle 5^n(5^2 - 5^3) = -2500$

    -Dan
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