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Thread: Can someone please do the first one log2 x showing working and I'll do the other two

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    Can someone please do the first one log2 x showing working and I'll do the other two


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    Re: Can someone please do the first one log2 x showing working and I'll do the other

    You haven’t learned how to differentiate log_a x (log x with base a). In order to do so, you need to use the ‘change of base’ rule to put the statement in appropriate form, with base/s ‘e’ (you should know how to differentiate logs with base e). Here, I’ll demonstrate:
    Log_a x = (log_e x)/(log_e a) [sorry If notation gets confusing]
    SO....
    Log_2 x = (ln x)/(ln 2) [remember log_e x = ln x]
    THEREFORE....
    THe differential of ‘log_2 x’ is the differential of ‘(ln x)/(ln 2)’, which is of course ‘1/xln2’

    Now you got the derivative, you can find your tangent gradient (let x=3) and then find the tangent (point-gradient formula).

    N.B. Remember with the ‘change of base’, you are able to create a fraction with a pair of logs with any base which is still equal to the original statement. Heck, we could’ve even used base 100 and still it would equal log_2 x; (log_100 x)/(log_100 2) = log_2 x. >>> but what’s the point of that? The base of ‘e’ is preferred since we can easily differentiate it using the small range of differentiation rules at our disposal.
    Last edited by tazhossain99; 20 Jan 2018 at 2:56 PM.
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    Re: Can someone please do the first one log2 x showing working and I'll do the other

    Quote Originally Posted by tazhossain99 View Post
    You haven’t learned how to differentiate log_a x (log x with base a). In order to do so, you need to use the ‘change of base’ rule to put the statement in appropriate form, with base/s ‘e’ (you should know how to differentiate logs with base e). Here, I’ll demonstrate:
    Log_a x = (log_e x)/(log_e a) [sorry If notation gets confusing]
    SO....
    Log_2 x = (ln x)/(ln 2) [remember log_e x = ln x]
    THEREFORE....
    THe differential of ‘log_2 x’ is the differential of ‘(ln x)/(ln 2)’, which is of course ‘1/xln2’

    Now you got the derivative, you can find your tangent gradient (let x=3) and then find the tangent (point-gradient formula).

    N.B. Remember with the ‘change of base’, you are able to create a fraction with a pair of logs with any base which is still equal to the original statement. Heck, we could’ve even used base 100 and still it would equal log_2 x; (log_100 x)/(log_100 2) = log_2 x. >>> but what’s the point of that? The base of ‘e’ is preferred since we can easily differentiate it using the small range of differentiation rules at our disposal.

    This is what I got can

    https://imgur.com/gallery/RWeQw

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    Senior Member integral95's Avatar
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    Re: Can someone please do the first one log2 x showing working and I'll do the other

    Quote Originally Posted by kpad5991 View Post
    This is what I got can

    https://imgur.com/gallery/RWeQw
    The change of base occurs again with the last term (changing from base 2 to base e), then you just factorise to get the given solution.
    “Smart people learn from their mistakes. But the real sharp ones learn from the mistakes of others.”
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    Re: Can someone please do the first one log2 x showing working and I'll do the other

    Quote Originally Posted by integral95 View Post
    The change of base occurs again with the last term (changing from base 2 to base e), then you just factorise to get the given solution.
    Can you show me these steps?

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    617 pages fan96's Avatar
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    Re: Can someone please do the first one log2 x showing working and I'll do the other






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