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# FP3 Hyperbolic Functions Question (edexcel) Watch

1. Hi everyone,

I'm somewhat confused over Q 9b. The result I got is right (a=-6) but given the way the Q is worded I reckon x should come to (1/6)ln1-(1/6)ln3 instead of what I got (giving -(1/6)ln3).

Can anyone verify whether I've made any mistakes? Solution attached. Thanks.
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2. (Original post by WatermelonJuice1)
Hi everyone,

I'm somewhat confused over Q 9b. The result I got is right (a=-6) but given the way the Q is worded I reckon x should come to (1/6)ln1-(1/6)ln3 instead of what I got (giving -(1/6)ln3).

Can anyone verify whether I've made any mistakes? Solution attached. Thanks.
do you know the inverse tanh formula?
3. (Original post by TeeEm)
do you know the inverse tanh formula?
Yep. Trying it now.
4. (Original post by WatermelonJuice1)
Yep. Trying it now.
i got tanh3x = -1/2
so x = -1/6ln3
5. (Original post by TeeEm)
i got tanh3x = -1/2
Got that bit as well but I'm not sure how to use the formula you mentioned to get the x. If tanh3x = -1/2 would I just say artanh(-1/2)=3x and so 3x=(1/2)ln[(1-1/2)/(1+1/2) ?
6. (Original post by WatermelonJuice1)
Got that bit as well but I'm not sure how to use the formula you mentioned to get the x. If tanh3x = -1/2 would I just say artanh(-1/2)=3x and so 3x=(1/2)ln[(1-1/2)/(1+1/2) ?
yes

so 3x = (1/2)ln(1/3)
so 3x = -(1/2)ln(3)
so x = -(1/6)ln(3)
7. (Original post by TeeEm)
yes

so 3x = (1/2)ln(1/3)
so 3x = -(1/2)ln(3)
so x = -(1/6)ln(3)
Ok I get it now. Thank you for your help kind sir.
8. (Original post by WatermelonJuice1)
Ok I get it now. Thank you for your help kind sir.
my pleasure

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