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Math Help - Solving exponential equations

  1. #1
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    Solving exponential equations

    3(2^(x+1)) + 2(5^(x-2)) = 5^x + 2^(x-2)

    I know I'm supposed to move the base 2s and base 5s to one side and factor out 2^x and 5^x, but I have no idea how to do it. Any help?
    Last edited by evaporate; June 19th 2010 at 02:25 PM. Reason: syntax
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  2. #2
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     3(2^{x+1}) + 2(5^{x-2}) = 5^x + 2^{x-2} rearranging we get

     3(2^{x+1}) +  2^{x-2}= 5^x +2(5^{x-2})  Now the laws of exponents tell us that  a^n + a^m = a^{m+n} so

     3(2^3)(2^{x-2}) + 2^{x-2} = (5^2)(5^{x-2}) +  2(5^{x-2}) (Note: It will almost always be easier to express in terms of the lowest power, in this case (x-2) for both exponents 2,5).

    Now factorise  2^{x-2} and  5^{x-2} then take the log of both sides. Use your log laws to solve for x.
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  3. #3
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    Quote Originally Posted by Gusbob View Post
    Now the laws of exponents tell us that  a^n + a^m = a^{m+n}
    Just in case there is still some confusion from this, Gusbob meant to say  a^m \times a^n = a^{m+n}
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  4. #4
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    Quote Originally Posted by Gusbob View Post
     3(2^{x+1}) + 2(5^{x-2}) = 5^x + 2^{x-2} rearranging we get

     3(2^{x+1}) +  2^{x-2}= 5^x +2(5^{x-2})
    No. We get  3(2^{x+1}) -  2^{x-2}= 5^x - 2(5^{x-2})
    Now the laws of exponents tell us that  a^n + a^m = a^{m+n} so
    (a^n)(a^m)= a^{m+n} which is used correctly below

     3(2^3)(2^{x-2}) + 2^{x-2} = (5^2)(5^{x-2}) +  2(5^{x-2}) (Note: It will almost always be easier to express in terms of the lowest power, in this case (x-2) for both exponents 2,5).

    Now factorise  2^{x-2} and  5^{x-2} then take the log of both sides. Use your log laws to solve for x.
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