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Need Help with ln(x)



Why is my method for part(a) wrong?

Because 'a' crosses the x axis, y must be 0. Therefore 0=ln(2+3x). I know that ln(1)=0 and so 2+3x=1 would give me solution for y=0. Therefore, my x value is -1/3... And this works perfectly.

In the mark scheme, it says x=-2/3 (which doesn't work)
Maths Q.png43.9KB
Reply 1
Avatar for BabyMaths
BabyMaths
Original post by ps1265A


Why is my method for part(a) wrong?

Because 'a' crosses the x axis, y must be 0. Therefore 0=ln(2+3x). I know that ln(1)=0 and so 2+3x=1 would give me solution for y=0. Therefore, my x value is -1/3... And this works perfectly.

In the mark scheme, it says x=-2/3 (which doesn't work)


Think again. :wink:

x=a is the asymptote that the curve 'approaches'.
(edited 9 years ago)
Reply 2
Avatar for james22
james22
16
Original post by ps1265A


Why is my method for part(a) wrong?

Because 'a' crosses the x axis, y must be 0. Therefore 0=ln(2+3x). I know that ln(1)=0 and so 2+3x=1 would give me solution for y=0. Therefore, my x value is -1/3... And this works perfectly.

In the mark scheme, it says x=-2/3 (which doesn't work)


The point a is not a point on the curve, you seem to think that it is.
Original post by ps1265A


Why is my method for part(a) wrong?

Because 'a' crosses the x axis, y must be 0. Therefore 0=ln(2+3x). I know that ln(1)=0 and so 2+3x=1 would give me solution for y=0. Therefore, my x value is -1/3... And this works perfectly.

In the mark scheme, it says x=-2/3 (which doesn't work)


You are correct that the curve crosses the axis at x=-1/3 but that is not the value of a.
a is the value of x such that ln(2+3x)=-infinity i.e. 2+3x=0.
Reply 4
Avatar for ps1265A
ps1265A
OP
14
Original post by BabyMaths
Think again. :wink:

x=a is the asymptote that the curve 'approaches'.


Thinking helps... thanks BabyMaths!

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