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    Can someone tell me what am I doing wrong? I keep getting wrong answer





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    (Original post by Moa)
    Can someone tell me what am I doing wrong? I keep getting wrong answer





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    If you are planning on studying at Imperial, C4 should be piss easy!
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    Where you say "c stays as a constant" and therefore add c (after exponentiating both sides), you should be multiplying by e^c = C, not adding.
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    Beaten to the punch. Yeah c = 4 and you're mutiplying not adding, as above stated
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    (Original post by Swayum)
    Where you say "c stays as a constant" and therefore add c (after exponentiating both sides), you should be multiplying by e^c = C, not adding.
    I do not see how it makes a difference I was always taught that:

    -as long as you do not perform any operations using unknowns (like dividing both sides by "x" or multiplying by "y" etc.) constant "c" stays the same.

    Here I do not perform any operations of this kind, I exponentiate both sides, however "c" is not in the exponentials power - it simply stays as +c, so why would I change it to e^c.

    The same way if I multiplied both sides by 2, "c" would stay as "c", not "2c"

    Can you tell me why am I wrong? Thank you for help
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    Rewrite your constant as lnA, where lnA is just another constant. Then when you are rewriting it the 2+y/2-y will be multiplied by A.
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    (Original post by Moa)
    I do not see how it makes a difference I was always taught that:

    -as long as you do not perform any operations using unknowns (like dividing both sides by "x" or multiplying by "y" etc.) constant "c" stays the same.

    Here I do not perform any operations of this kind, I exponentiate both sides, however "c" is not in the exponentials power - it simply stays as +c, so why would I change it to e^c.

    The same way if I multiplied both sides by 2, "c" would stay as "c", not "2c"

    Can you tell me why am I wrong? Thank you for help
    The same reason for when you multiplied 1/2-y by 2+y when you brought it down from the exponential. The e^c is also multiplied.
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    By the laws of logarithims y=x+a does not imply lny=lnx+a, if you like.
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    (Original post by Moa)
    Here I do not perform any operations of this kind, I exponentiate both sides, however "c" is not in the exponentials power - it simply stays as +c, so why would I change it to e^c.
    It is in the power though.

    a = b + c

    Goes to e^a = e^(b+c) - you need to exponentiate both sides and the RHS includes c.

    And if you say let e^c = d, then you get e^a = de^b.

    Here a = your LHS and b = those logarithms.

    The same way if I multiplied both sides by 2, "c" would stay as "c", not "2c"
    That's just bad notation for convenience. Writing 2c as c works because it's like saying "Well, when I integrated, I could have added c/2 instead of c and so multiplying by 2 would give me c on this line, but I can't be bothered to cross out my previous working, so I'll continue calling it c".
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    Right, I guess from now on I will always properly write what happens with c Thanks guys!
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    Going back to my a = b + c example:

    Let's say we want y = x + c, where c is a constant.

    Your claim is that e^y = e^(x) + c, so c = e^y - e^x.

    Now let's say my line y = x + c passes through (3, 5), then c = 2 right?

    Then c = 2 = e^y - e^x, but 2 = e^5 - e^3 isn't true...
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    (Original post by Swayum)
    Going back to my a = b + c example:

    Let's say we want y = x + c, where c is a constant.

    Your claim is that e^y = e^(x) + c, so c = e^y - e^x.

    Now let's say my line y = x + c passes through (3, 5), then c = 2 right?

    Then c = 2 = e^y - e^x, but 2 = e^5 - e^3 isn't true...
    Thanks! That clears everything up
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    this isnt edexcel is it ?
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    (Original post by bonbon)
    this isnt edexcel is it ?
    it is
 
 
 
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