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.

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