FP3 Hyperbolic Functions Question (edexcel)

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?

Yep. Trying it now.

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) ?

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.