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Math Help - Solving equation using equality rule for exponents

  1. #1
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    Solving equation using equality rule for exponents

    {5^(2x-2x^2)}={24*5^-2}+{5^(-2x2)}

    Solve for X
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  2. #2
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    Quote Originally Posted by etle1989 View Post
    {5^(2x-2x^2)}={24*5^-2}+{5^(-2x2)}

    Solve for X
    Please fix the formatting so that the equation is clear and unambiguous. Note: Code for things like 5^{2x} is 5^{2x}, NOT 5^2x. Failure to use the correct code will result in things like 5^2x.
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  3. #3
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    Quote Originally Posted by etle1989 View Post
    5^{2x-2x^2}=24*5^{-2}+5^{-2x^2}

    Solve for X
    That's my guess at what you really mean. That " 24*5^{-2}" really complicates things since we do NOT have just " 5^a= 5^b

    We can rewrite this as 5^{2x- 2x^2}- 5^{-2x^2}= 24*5^{-2} which is the same as 5^{2x}5^{-2x2}- 5^{2x^2}= (5^{2x}- 1)(5^{-2x^2})= 24*5^{-2}. Now, I note that if x= 1, then I have 5^{2x}-1= 25- 1= 24 and 5^{-2x^2}= 5^{-2}. However, in general, you cannot just assume that if ab= 24*5^{-2} then a= 24 and b= 5^{-2}. We can see that x= 1 is a solution but we do not yet know if there are other solutions.
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  4. #4
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    Quote Originally Posted by etle1989 View Post
    5^{2x-2x^2}=24*5^{-2}+5^{-2x^2}

    Solve for X
    1. Re-write the equation:

    \dfrac{5^{2x}}{5^{2x^2}} = \dfrac{24}{25} + \dfrac{1}{5^{2x^2}}

    2. Collecting the variables at the LHS:

    \dfrac{5^{2x}}{5^{2x^2}} -  \dfrac{1}{5^{2x^2}}= \dfrac{24}{25}

    \dfrac{5^{2x}-1}{5^{2x^2}} = \dfrac{24}{25}

    3. That means you have to solve:

    5^{2x}-1 = 24~\wedge~5^{2x^2} = 25

    5^{2x} = 25 = 5^2~\wedge~5^{2x^2} = 25 = 5^2

    2x = 2~\wedge~2x^2=2

    which yields x = 1 as the only solution.
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  5. #5
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    1 isn't the only solution.

    There are 2 solutions:

    \displaystyle x = 1

    \displaystyle x \approx 0.238129
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  6. #6
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    Quote Originally Posted by Educated View Post
    1 isn't the only solution.

    There are 2 solutions:

    \displaystyle x = 1

    \displaystyle x \approx 0.238129
    Thank you for pointing this out.
    Unfortunately I was only able to get a approximate numerical solution:

    5^{2x-2x^2}=24*5^{-2}+5^{-2x^2}

    5^{2x}=\frac{24}{25}\cdot 5^{2x^2}+1

    5^{2x}=24\cdot 5^{2x^2-2}+1

    0=24\cdot 5^{2x^2-2}-5^{2x}+1

    Now using Nerwton's method yields x \approx 0.2381291343
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  7. #7
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    Quote Originally Posted by earboth View Post
    1. Re-write the equation:

    \dfrac{5^{2x}}{5^{2x^2}} = \dfrac{24}{25} + \dfrac{1}{5^{2x^2}}

    2. Collecting the variables at the LHS:

    \dfrac{5^{2x}}{5^{2x^2}} -  \dfrac{1}{5^{2x^2}}= \dfrac{24}{25}

    \dfrac{5^{2x}-1}{5^{2x^2}} = \dfrac{24}{25}

    3. That means you have to solve:

    5^{2x}-1 = 24~\wedge~5^{2x^2} = 25
    this does not follow The fact that \frac{x}{y}= \frac{a}{b} does NOT imply that x= a and y= b!
    It does happen to be true for x= 1 but it is not true that we must separate the fractions that way.

    5^{2x} = 25 = 5^2~\wedge~5^{2x^2} = 25 = 5^2

    2x = 2~\wedge~2x^2=2

    which yields x = 1 as the only solution.
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  8. #8
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    5^{2x}=24\cdot 5^{2x^2-2}+1

    still dont understand how i can simplify that to use equality rule. can someone please explain.
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  9. #9
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    Quote Originally Posted by etle1989 View Post
    5^{2x}=24\cdot 5^{2x^2-2}+1

    still dont understand how i can simplify that to use equality rule. can someone please explain.
    What you want to do cannot be done. The posts in this thread have shown you how the question needs to be solved.
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