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    Stuck on Q2 C3 Soloman Paper I, b)
    What is the log of (-2e^x) .. (-2x)? Think this is where I went wrong.
    Thanks
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    -2ex is negative, so cannot have a logarithm ?

    :beard:
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    (Original post by ColeWorld98)
    Stuck on Q2 C3 Soloman Paper I, b)
    What is the log of (-2e^x) .. (-2x)? Think this is where I went wrong.
    Thanks
    You have y=3+2e^x and  y=e^{x+2}, so at the point of intersection, e^{x+2} = 3+2e^x, can you see that you can write e^{x+2} as e^x\cdot e^2
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    (Original post by the bear)
    -2ex is negative, so cannot have a logarithm ?

    :beard:
    (Original post by NotNotBatman)
    You have y=3+2e^x and  y=e^{x+2}, so at the point of intersection, e^{x+2} = 3+2e^x, can you see that you can write e^{x+2} as e^x\cdot e^2
    (Original post by the bear)
    -2ex is negative, so cannot have a logarithm ?

    :beard:
    (Original post by NotNotBatman)
    You have y=3+2e^x and  y=e^{x+2}, so at the point of intersection, e^{x+2} = 3+2e^x, can you see that you can write e^{x+2} as e^x\cdot e^2
    Hi guys
    Yes i noticed that but that was only on realising that my first method did go wrong somewhere as the X co ordinate was + when graph clearly showed to be -ve, can you check the attachment - i cant see why my first method didnt work ?
    Thanks for all ur help
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    could you also post the question for us ?

    :holmes:
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    (Original post by ColeWorld98)
    Hi guys
    Yes i noticed that but that was only on realising that my first method did go wrong somewhere as the X co ordinate was + when graph clearly showed to be -ve, can you check the attachment - i cant see why my first method didnt work ?
    Thanks for all ur help
    Third line of part b, you have to take logs to the entire RHS if you're going to do that. It would be on the rhs, ln(e^{x+2} - 2e^x)

    Which isn't just the part in the brackets.
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    (Original post by NotNotBatman)
    Third line of part b, you have to take logs to the entire RHS if you're going to do that. It would be on the rhs, ln(e^{x+2} - 2e^x)

    Which isn't just the part in the brackets.
    But how would you simplify
    Just cant see it
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    (Original post by ColeWorld98)
    But how would you simplify
    Just cant see it
    You don't, remember = , you would factorise e^x from line 2 and solve from there.
 
 
 
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