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Math Help - exponential equation

  1. #1
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    exponential equation

    9^(x+1) + 3^(x+2) - 4 = 0

    can anyone show me how to solve this.
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  2. #2
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    Quote Originally Posted by fabregas99 View Post
    9^(x+1) + 3^(x+2) - 4 = 0

    can anyone show me how to solve this.
    Since 9^{x+1} = 9^x \cdot 9 and 9 = 3^2 it follows that 9^{x+1} = 9 \cdot (3^2)^x = 9 \cdot (3^x)^2.

    Furthermore, 3^{x+2} = 3^x \cdot 3^2 = 9 \cdot 3^x.

    Therefore your equation can be re-written as

    9 \cdot (3^x)^2 + 9 \cdot 3^x - 4 = 0 \Rightarrow (3 \cdot 3^x + 4) (3 \cdot 3^x - 1) = 0.
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  3. #3
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    thanks

    i've tried working it out, but i cant get the answer that it has in the back of the practice exam. which is

    x=log3(4/3)

    (thats log to the base 3 couldnt find subscript)
    9^(x+1) = 3^(x+3)
    so
    3^(x+3) - 3^(x+2) -4 = 0

    then you can take logs to the base 3 of the first 2 exponentials but when it come to 4, i relaly dont understand what to do. reall sorry on my first post just realised i put a + when it should have been a - imbetween 9^(x+1) and 3^(x+2). can u explain.
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  4. #4
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    Quote Originally Posted by fabregas99 View Post
    i've tried working it out, but i cant get the answer that it has in the back of the practice exam. which is

    x=log3(4/3)

    (thats log to the base 3 couldnt find subscript)
    9^(x+1) = 3^(x+3)
    so
    3^(x+3) - 3^(x+2) -4 = 0

    then you can take logs to the base 3 of the first 2 exponentials but when it come to 4, i relaly dont understand what to do. reall sorry on my first post just realised i put a + when it should have been a - imbetween 9^(x+1) and 3^(x+2). can u explain.
    Then you make a small change to what I posted - the equation can be re-written as

    9 \cdot (3^x)^2 - 9 \cdot 3^x - 4 = 0 \Rightarrow (3 \cdot 3^x - 4) (3 \cdot 3^x + 1) = 0.

    So one of the two bracketed expressions is equal to zero ....
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  5. #5
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    Quote Originally Posted by fabregas99 View Post
    9^(x+1) + 3^(x+2) - 4 = 0
    Use the rules for exponents to restate in the manner outlined in a previous reply:

    . . . . . 9^{x+1}\, =\, 9^x\, \times\, 9^1\, =\, (9^x)(9)\, =\, 9\, \times\, 9^x

    . . . . . 3^{x+2}\, =\, 3^x\, \times\, 3^2\, =\, 3^x\, \times\, 9\, \, 9\, \times\, 3^x

    This makes the equation the same as:

    . . . . . 9(9^x)\, +\, 9(3^x)\, -\, 4\, =\, 0

    Then note that 9 = 3^2, so 9^x = (3^2)^x = (3^x)^2, which means the equation is really a "quadratic" in 3^x:

    . . . . . 9(3^x)^2\, +\, 9(3^x)\, -\, 4\, =\, 0

    Using the factoring provided earlier, you should arrive at:

    . . . . . 3^x\, =\, -\frac{4}{3}\, \mbox{ and }\, 3^x\, =\, \frac{1}{3}

    From what you've learned about exponential functions, you know that you cannot, by powers, turn a positive three into a negative value, so the first of the two equations above has "no solution".

    Apply logs or the definition to solve the other exponential equation.

    Note: The solution provided cannot be correct. If x = log_3(4/3), then log definitions and log rules give::

    . . . . . x\, +\, 1\, =\, \log_3(4/3)\, +\, \log_3(3)\, =\, \log_3(4)

    . . . . . x\, +\, 2\, =\, \log_3(4/3)\, +\, \log_3(9)\, =\, \log_3(8)

    . . . . . 9^{x+1}\, =\, \left(3^{\log_3(4)}\right)^2\, =\, (4)^2\, =\, 16

    . . . . . 3^{x+2}\, =\, 3^{\log_3(8)}\, =\, 8

    . . . . . 9^{x+1}\, +\, 3^{x+2}\, -\, 4\, =\, 16\, +\, 8\, -\, 4\, =\, 20

    But the left-hand side is supposed to equal zero.

    You might want to point out this typo to your instructor.
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  6. #6
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    Quote Originally Posted by stapel View Post
    Use the rules for exponents to restate in the manner outlined in a previous reply:

    . . . . . 9^{x+1}\, =\, 9^x\, \times\, 9^1\, =\, (9^x)(9)\, =\, 9\, \times\, 9^x

    . . . . . 3^{x+2}\, =\, 3^x\, \times\, 3^2\, =\, 3^x\, \times\, 9\, \, 9\, \times\, 3^x

    This makes the equation the same as:

    . . . . . 9(9^x)\, +\, 9(3^x)\, -\, 4\, =\, 0

    Then note that 9 = 3^2, so 9^x = (3^2)^x = (3^x)^2, which means the equation is really a "quadratic" in 3^x:

    . . . . . 9(3^x)^2\, +\, 9(3^x)\, -\, 4\, =\, 0

    Using the factoring provided earlier, you should arrive at:

    . . . . . 3^x\, =\, -\frac{4}{3}\, \mbox{ and }\, 3^x\, =\, \frac{1}{3}

    From what you've learned about exponential functions, you know that you cannot, by powers, turn a positive three into a negative value, so the first of the two equations above has "no solution".

    Apply logs or the definition to solve the other exponential equation.

    Note: The solution provided cannot be correct. If x = log_3(4/3), then log definitions and log rules give::

    . . . . . x\, +\, 1\, =\, \log_3(4/3)\, +\, \log_3(3)\, =\, \log_3(4)

    . . . . . x\, +\, 2\, =\, \log_3(4/3)\, +\, \log_3(9)\, =\, \log_3(8)

    . . . . . 9^{x+1}\, =\, \left(3^{\log_3(4)}\right)^2\, =\, (4)^2\, =\, 16

    . . . . . 3^{x+2}\, =\, 3^{\log_3(8)}\, =\, 8

    . . . . . 9^{x+1}\, +\, 3^{x+2}\, -\, 4\, =\, 16\, +\, 8\, -\, 4\, =\, 20

    But the left-hand side is supposed to equal zero.

    You might want to point out this typo to your instructor.
    Memo to stapel:

    Quote Originally Posted by fabregas99 View Post
    [snip]reall sorry on my first post just realised i put a + when it should have been a - imbetween 9^(x+1) and 3^(x+2).
    [snip]
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