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    Any help solving this would be immensely appreciated x
    2^(2x+3)= 3^(3x+2)
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    (Original post by Hardy jacks)
    Any help solving this would be immensely appreciated x
    2^(2x+3)= 3^(3x+2)
    First start by taking logs of both sides so you get
     \log{2^{(2x+3)}} = \log{3^{(3x+2)}}

    How could you then simplify that?
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    (Original post by KaylaB)
    First start by taking logs of both sides so you get
     \log{2^{(2x+3)}} = \log{3^{(3x+2)}}

    How could you then simplify that?


    Well applying change of base straight away didn't work
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    (Original post by Hardy jacks)
    Well applying change of base straight away didn't work
    When there's a power after logging, you can bring it down to the left, then it's just algebra manipulation
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    log{a^b}={b} log{a}
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    Alternatively, you should get the same result by using 3=2^{log_2{3}}=2^\frac{log3}{log  2}
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    (Original post by Hardy jacks)
    Well applying change of base straight away didn't work
    Apologies,  \log2 = \log_{10}2 It's so useful that we just shorten it!
    You could absolutely take logs with whatever base you like for these, it doesn't really matter.

    In terms of simplifying  \log_{10}{2^{(2x+3)}} = \log_{10}{3^{(3x+2)}}
    Are you familiar with this rule?
     \log_{b}{x^{y}} = y\log_{b}{x}

    Even so, how could that rule be applied to the above so that we can simplify it and solve for x?
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    [QUOTE=RogerOxon;69638674]Alternatively, you should get the same result by using 3=2^{log_2{3}}=2^\frac{log3}{log  2}[/QUOTE
    I'm sorry I'm really confused could anyone actually get -0.06... as their answer?
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    take log of both sides -> use log(a^b) = b x log(a) -> expand brackets -> rearrange for x
    if you've done the manipulation correctly you'll get this:
    x = log(9/8) / log(4/27)
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    (Original post by Hardy jacks)
    I'm sorry I'm really confused could anyone actually get -0.06... as their answer?
    Your answer (-0.0617 to 3 s.f.) looks correct - putting it into the original equation gives:

    2^{2x+3}=2^{2.877}=7.34

    3^{3x+2}=3^{1.815}=7.34

    Here's my full working:

    2^{2x+3}=3^{3x+2}

    Take log of each side:
    log{2^{2x+3}}=log{3^{3x+2}}
    (2x+3)log2=(3x+2)log3
    x(2log2-3log3)=2log3-3log2
    x(log{2^2}-log{3^3})=log{3^2}-log{2^3}
    xlog{\frac{4}{27}}=log{\frac{9}{  8}}
    \therefore x=\frac{log{\frac{9}{8}}}{log{ \frac{4}{27}}}

    The other approach is:
    2^{2x+3}=3^{3x+2}={2^(\frac{log3  }{log2}})^{3x+2}=2^\frac{(3x+2)l  og3}{log2}
    \therefore 2x+3=\frac{log3(3x+2)}{log2}
    \therefore (2x+3)log2=(3x+2)log3
    Which is the same as above.
 
 
 
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