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    How would I answer 3 x 2^x = 7^x-2. I keep trying and getting -29.145, which is obviously wrong. Thanks in advanceName:  IMG_5947.jpg
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    (Original post by doglover123)
    How would I answer 3 x 2^x = 7^x-2. I keep trying and getting -29.145, which is obviously wrong. Thanks in advanceName:  IMG_5947.jpg
Views: 94
Size:  407.2 KB
    I'm sorry the photo is sideways, i have no idea how to rotate it
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    Bring the x terms to one side and factorise.
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    (Original post by doglover123)
    How would I answer 3 x 2^x = 7^x-2. I keep trying and getting -29.145, which is obviously wrong. Thanks in advanceName:  IMG_5947.jpg
Views: 94
Size:  407.2 KB
    What I done was divide both sides by 2^{x}

    So 3 = \dfrac{7^{x-2}}{2^{x}}

    Then take logs of both sides and use the log(\frac{x}{y}) = logx - logy rule. Then find a way of factorising x

    Your mistake lies in the first line of your working out. Taking log to 3(2^{x}) does not equal 3log(2^{x})

    It is log[3(2^{x})] and again from here you can use the log(xy) = logx + logy rule to work it out also
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    (Original post by edothero)
    What I done was divide both sides by 2^{x}

    So 3 = \dfrac{7^{x-2}}{2^{x}}

    Then take logs of both sides and use the log(\frac{x}{y}) = logx - logy rule. Then find a way of factorising x

    Your mistake lies in the first line of your working out. Taking log to 3(2^{x}) does not equal 3log(2^{x})

    It is log[3(2^{x})] and again from here you can use the log(xy) = logx + logy rule to work it out also
    THANK YOU SO MUCH! Everything became a lot clearer once I divided by 2^x
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    (Original post by doglover123)
    THANK YOU SO MUCH! Everything became a lot clearer once I divided by 2^x
    No problem I'm glad I helped. Also I hope you took note about your mistake and why it was wrong! Always good to learn from mistakes
 
 
 
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