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Can someone please do the first one log2 x showing working and I'll do the other two (1 Viewer)
- Thread starter kpad5991
- Start date
tazhossain99
Member
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.
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:
Re: Can someone please do the first one log2 x showing working and I'll do the other
This is what I got can
https://imgur.com/gallery/RWeQw
This is what I got can
https://imgur.com/gallery/RWeQw
integral95
Well-Known Member
- Joined
- Dec 16, 2012
- Messages
- 779
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- Male
- HSC
- 2013
Re: Can someone please do the first one log2 x showing working and I'll do the other
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.
Re: Can someone please do the first one log2 x showing working and I'll do the other
Can you show me these steps?
