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Math Help - Solve this exponential equation

  1. #1
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    Solve this exponential equation

    If someone could help me to solve this, I would appreciate it a lot.
    Result too, please.

    (3/5)^x-1 * (2/3)^x = 40/225
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Shandy View Post
    If someone could help me to solve this, I would appreciate it a lot.
    Result too, please.

    (3/5)^x-1 * (2/3)^x = 40/225
    Here's a start

    \left( \frac 35 \right)^{x - 1} \cdot \left( \frac 23 \right)^x = \frac {40}{225}

    \Rightarrow \left( \frac 35 \right)^{-1} \cdot \left( \frac 35 \right)^x \cdot \left( \frac 23 \right)^x = \frac {40}{225}

    \Rightarrow \frac 53 \cdot \left( \frac 35 \cdot \frac 23 \right)^x = \frac {40}{225}

    \Rightarrow \left( \frac 25 \right)^x = \frac 8{75}
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  3. #3
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    Hello Shandy
    Quote Originally Posted by Shandy View Post
    If someone could help me to solve this, I would appreciate it a lot.
    Result too, please.

    (3/5)^x-1 * (2/3)^x = 40/225
    \left(\frac{3}{5}\right)^{x-1}\times\left(\frac{2}{3}\right)^{x} = \frac{40}{225}

    Take logs of both sides, and use the laws:

    \log(a^b) = b\log (a) and \log(a\times b) = \log(a) + \log(b)

    and get:

    (x-1) \log \left(\frac{3}{5}\right) + x\log\left(\frac{2}{3}\right) = \log\left(\frac{40}{225}\right)

    Collect together the like terms:

    x\left( \log \left(\frac{3}{5}\right) + \log\left(\frac{2}{3}\right)\right) = \log\left(\frac{40}{225}\right)+ \log \left(\frac{3}{5}\right)

    \Rightarrow x\log\left(\frac{3\times 2}{5 \times 3}\right) = \log\left(\frac{40\times 3}{225 \times 5} \right)

    \Rightarrow x\log\left(\frac{2}{5}\right) = \log\left(\frac{8}{75}\right)

    \Rightarrow x = 2.4425

    Grandad
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