The Student Room Group

The function lnx

Is this right?

e^2x+1 = 2
2x +1 = ln 2
x = 0.5 ln( 2- 1)

I though for the last step:

x = 0.5 (ln(2) -1)

with the one outside the ln brakets?
You are correct. The last step is x = 0.5 (ln(2)-1)
Reply 2
Original post by CharlieBoardman
You are correct. The last step is x = 0.5 (ln(2)-1)


Thats what is thought but the mark sheme says this

0.5 ln( 2- 1)
Reply 3
Original post by nazgul60
Thats what is thought but the mark sheme says this

0.5 ln( 2- 1)


That's wrong.
Reply 4
Original post by nazgul60
Thats what is thought but the mark sheme says this

0.5 ln( 2- 1)


It is wrong because that gives 0.5ln(1) = 0 and this can be easily disproved by substituting x=0 into the initial equation.
Reply 5
Original post by Micky76
It is wrong because that gives 0.5ln(1) = 0 and this can be easily disproved by substituting x=0 into the initial equation.


Thats what i thought but just needed to be confident. Thanks!
Reply 6
Am I just being stupid or isn't x=0 the answer?
Reply 7
e2x+1=2e^{2x+1}=2

or

e2x+1=2e^{2x}+1=2
Reply 8
Original post by BabyMaths
e2x+1=2e^{2x+1}=2

or

e2x+1=2e^{2x}+1=2


Is it not multiplied by e instead of the +1?
Reply 9
Original post by BabyMaths
e2x+1=2e^{2x+1}=2





that one
Reply 10
Original post by BabyMaths
e2x+1=2e^{2x+1}=2

or

e2x+1=2e^{2x}+1=2


Sorry, my bad in the interpretation of your post.
Reply 11
Original post by nazgul60
that one


In that case x=12(1+ln2)x=\frac{1}{2}(-1+\ln 2) as you thought.
Original post by nazgul60
Thats what is thought but the mark sheme says this

0.5 ln( 2- 1)

Sometimes mark schemes have mistakes in them. Don't worry - you are correct! :smile:
Reply 13
Original post by nazgul60
that one


Then you are right/mark scheme is wrong.

probably just a typing error by a clerk who doesn't know to be careful with these things.

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