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Math Help - Exponents and Logs

  1. #1
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    Exponents and Logs

    Why is the 5 in this equation, I can't multiply the 2 and 5, some of this math is so confusing

    2(5)^x=7^{x+1}
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  2. #2
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    Quote Originally Posted by nuckers View Post
    Why is the 5 in this equation, I can't multiply the 2 and 5, some of this math is so confusing

    2(5)^x=7^{x+1}
    The 5 is the base of the exponential. You can still take logs and use the relevant log laws.

    Hint: \log(2\cdot 5^x) = \log(2) + \log(5^x)
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  3. #3
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    Quote Originally Posted by nuckers View Post
    2(5)^x=7^{x+1}
    That equation is equivalent to \left( {\frac{5}{7}} \right)^x  = \frac{7}{2} which has solution x = \frac{{\log (7) - \log (2)}}{{\log (5) - \log (7)}}
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  4. #4
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    Quote Originally Posted by Plato View Post
    That equation is equivalent to \left( {\frac{5}{7}} \right)^x = \frac{7}{2} which has solution x = \frac{{\log (7) - \log (2)}}{{\log (5) - \log (7)}}
    I'm totally confused how you came up with that. I did screw up and put a 7 in the equation, it's supposed to be 3^{x+1}
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  5. #5
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    Quote Originally Posted by nuckers View Post
    I'm totally confused how you came up with that. I did screw up and put a 7 in the equation, it's supposed to be 3^{x+1}
     2 \cdot 5^x  = 3^{x + 1} is equivalent to \left( {\frac{5}{3}} \right)^x  = \frac{3}{2} which has solution x = \frac{{\log (3) - \log (2)}}{{\log (5) - \log (3)}}
    To see how that works divide both sides by 3^x and also by 2.
    Note that \frac{{5^x }}{{3^x }} = \left( {\frac{5}{3}} \right)^x .
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  6. #6
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    Quote Originally Posted by Plato View Post
     2 \cdot 5^x = 3^{x + 1} is equivalent to \left( {\frac{5}{3}} \right)^x = \frac{3}{2} which has solution x = \frac{{\log (3) - \log (2)}}{{\log (5) - \log (3)}}
    To see how that works divide both sides by 3^x and also by 2.
    Note that \frac{{5^x }}{{3^x }} = \left( {\frac{5}{3}} \right)^x .
    Ok, i understand it now, thanks so much for the help, i appreciate it
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