If #H(x)=6ln(x^2-5)+3ln(4)#, what is #H^-1(x)#?

Question

1068 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-31T16:00:09

#g(x) = sqrt(1/2e^(1/6x)+5)#

The domain where the function is bijective is #I = ]sqrt(5):oo[#

We restrict the study on this domain

#y = 6ln(x^2-5)+3ln(4)#

Don't forget #aln(b) = ln(b^a)#

#y = 6ln(x^2-5)+6ln(2)#

#y = 6(ln(x^2-5)+ln(2))#

We just need to solve for #x#

#1/6y=ln(x^2-5)+ln(2)#

#1/6y-ln(2)=ln(x^2-5)#

Take exponential both side

#e^(1/6y-ln(2))=x^2-5#

#1/2e^(1/6y)+5=x^2#

#sqrt(1/2e^(1/6y)+5)=x#