The Student Room Group
Also another graph question:

4 - x^2 = (modulas) 2x-1

how are we meant to find the exaclt solutions???
Well if y=lnx y = lnx then x=ey x = e^y . Does that help?
DreamsComeTrue.
given that y = ln x, find expressions in terms of y for

(i) log (base 2) x

(ii)lnx2/e ln {x^2}/{e}

help me please!! i hated this since C2 :frown:

logax=logbxlogba\log_a{x}=\frac{\log_b{x}}{\log_b{a}}

sort your latex out for the second one.
DreamsComeTrue.
Also another graph question:

4 - x^2 = (modulas) 2x-1

how are we meant to find the exaclt solutions???

draw a graph.
jonny23563
Well if y=lnx y = lnx then x=ey x = e^y . Does that help?


yeah it does! i complety forgot about doing that! thanks :smile:
DreamsComeTrue.
Also another graph question:

4 - x^2 = (modulas) 2x-1

how are we meant to find the exaclt solutions???


To solve 4x2=2x1 4 -x^2 = |2x -1| you need to remember that f(x)=f(x)f(x) = -f(x) for x<0 x< 0 and so swap the signs of the RHS accordingly.I think personally graph-drawing is a little unnecessary here.

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