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    4y = 15ln (2) x - 15 ln(2) 2 + 17

    At the back of the book , they have took out as a common factor the 15 ln(2) in order to simplify stuff, but then i thought that it couldnt be taken out since it doesnt appear in all three terms.

    Thanks for all the help in advance!
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    (Original post by laurawoods)
    4y = 15ln (2) x - 15 ln(2) 2 + 17

    At the back of the book , they have took out as a common factor the 15 ln(2) in order to simplify stuff, but then i thought that it couldnt be taken out since it doesnt appear in all three terms.

    Thanks for all the help in advance!
    Can you post the initial question?
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    (Original post by laurawoods)
    4y = 15ln (2) x - 15 ln(2) 2 + 17

    At the back of the book , they have took out as a common factor the 15 ln(2) in order to simplify stuff, but then i thought that it couldnt be taken out since it doesnt appear in all three terms.

    Thanks for all the help in advance!
    Can you post the original question

    As you say ln2 is not a common factor so perhaps the error was earlier in your work
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    (Original post by notnek)
    Can you post the initial question?
    hello there,
    The initial question was

    Find the equation of the tangent to the curve y= 2^x +2^-x at the point (2, 17/4).

    thanku for all help in advance!
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    What did they get as a result? I don't see how factorizing can "simplify" an equation in the first place.

    A move from

    4y = 15\ln 2 (x) - 15 \ln 2 (2) + 17

    to

    4y = 15\ln 2 (x - 2)  + 17

    Is perfectly valid...
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    (Original post by aznkid66)
    What did they get as a result? I don't see how factorizing can "simplify" an equation in the first place.

    A move from

    4y = 15\ln 2 (x) - 15 \ln 2 (2) + 17

    to

    4y = 15\ln 2 (x - 2)  + 17

    Is perfectly valid...
    Hello , when determining the nature of any stationary points , if the second derivative is equal to 0, we differentiate again, right? and then for it to be a point of inflection what must the third differential equal to? thanks
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    The third derivative must exist and not be 0, for this implies that there is a sign change at the second derivative.
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    (Original post by laurawoods)
    Hello , when determining the nature of any stationary points , if the second derivative is equal to 0, we differentiate again, right? and then for it to be a point of inflection what must the third differential equal to? thanks
    For a point of inflection, the third derivative must be non-zero.
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    (Original post by aznkid66)
    The third derivative must exist and not be 0, for this implies that there is a sign change at the second derivative.
    Where do you keep getting this "third derivative" thing from? It's got nothing to do with a point of inflexion.

    If the 2nd derivative is 0 then you may or may not have a point of inflexion. If the 2nd derivative is -ve on one side of the point and +ve on the other, then that point is a "point of inflexion" (the curvature changes sign).

    (The first derivative can be 0 at a point of inflexion, but it MUST NOT change sign at that point - if it does, then you have a max or min.)
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    (Original post by Indeterminate)
    For a point of inflection, the third derivative must be non-zero.
    Er, y=x^5, x^7, x^9 etc all have points of inflexion at x=0. The third derivative is irrelevant.
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    (Original post by aznkid66)
    The third derivative must exist and not be 0, for this implies that there is a sign change at the second derivative.
    Hello there, thanks !

    Pls can u help me with understading this:
    why cant 4ln(2) - 2 be written as ln(2)^4 - 2 ?
    when i type into calc they are giving different values..
    thanks, laurawoods.
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    Make sure you are doing ln(2^4) and not [ln(2)]^4
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    (Original post by aznkid66)
    Make sure you are doing ln(2^4) and not [ln(2)]^4
    hello there, thanks ...i didnt know it had to be inside...so it is it the number inside raised to k , and not the whole thing being raised to k , right?

    thanks
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    Yupyup, that is what the identity states

    You can also get the identity from combining n*ln(x)=ln(x)+ln(x)+...+ln(x)=ln (x*x*...*x)=ln(x^n)
 
 
 
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