How do i find or approximate the points that are guaranteed to exist by the mean value theorem on f(x)=ln2x; [1,e] ?

Question

How do i find or approximate the points that are guaranteed to exist by the mean value theorem on f(x)=ln2x; [1,e]  ?

anonymous · Answer

$$f'(c)=\frac{f(b)-f(a)}{b-a}$$
$$a=1, b=e, f(1)=\ln(2), f(e)=\ln(2e)=\ln(2)+\ln(e)=\ln(2)+1$$
$$f'(x)=\frac{1}{x}$$

anonymous · Answer

so your job is to solve for c
$$\frac{1}{c}=\frac{\ln(2)+1-\ln(2)}{e-1}$$
$$\frac{1}{c}=\frac{1}{e-1}$$

anonymous · Answer

alright, i'll get on that

anonymous · Answer

one step left! take the reciprocal and you are done

anonymous · Answer

okay, i'll piece it together from here, thanks

anonymous · Answer

i just gotta look it all over

anonymous · Answer

the statement of the mvt is easy to read, the hard part is in going from the general 
$$f'(c)=\frac{f(b)-f(a)}{b-a}$$ to the specific

anonymous · Answer

so it would just be c=e-1?

anonymous · Answer

or c=(c^2)/(e-1)

anonymous · Answer

i got $$c=e-1$$