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Math Help - solving exponents with logs

  1. #1
    Super Member bigwave's Avatar
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    solving exponents with logs

    Given:
     <br />
5^{x+3} - 5^{x+2} = 2^{3x}<br />
    Find: x
    Answer is: 9.8

    here is my steps

     (x+3)\log{5} - (x+2)\log{5} = (3x)\log{2}
     \log{5}[(x+3)-(x+2)] = (3x)\log{2}
    \frac{\log{5}}{\log{2}} = 3x
     2.322 = 3x
    .774 = x
    however this is not the answer.
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  2. #2
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    Hello, bigwave!

    Given: .  5^{x+3} - 5^{x+2} \:=\: 2^{3x}
    Find x.

    Answer: 9.8

    Here is my steps

     (x+3)\log{5} - (x+2)\log{5} = (3x)\log{2} . . . . wrong!

    We must perform some gymnastics first . . .


    \begin{array}{ccccc}\text{We have:} & 5^{x+3} - 5^{x+2} &=& 2^{3x} \\ \\ \text{Factor:} & 5\cdot5^{x+2} - 5^{x+2} &=& 2^{3x} \\ \\ \text{Factor:} & (5-1)5^{x+2} &=& 2^{3x} \\ \\ \text{We have:} & 4\cdot5^{x+2} &=& 2^{3x} \\ \\ \text{Divide by 4:} & 5^{x+2} &=& 2^{3x-2}\end{array}



    Now take logs: . \ln\left(5^{x+2}\right) \;=\;\ln\left(2^{3x-2}\right) \quad\Rightarrow\quad (x+2)\!\cdot\!\ln(5) \;=\;(3x-2)\!\cdot\!\ln(2)

    . . and so on . . .

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  3. #3
    Super Member bigwave's Avatar
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    Quote Originally Posted by Soroban View Post
    Now take logs: . \ln\left(5^{x+2}\right) \;=\;\ln\left(2^{3x-2}\right) \quad\Rightarrow\quad (x+2)\!\cdot\!\ln(5) \;=\;(3x-2)\!\cdot\!\ln(2)
    as suggested...

     (x+2)(1.609)=(3x-2)(.693)
     1.609x+3.218=2.079x-1.386
    -.470x = -4.604
    x = 9.8
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