0580 Maths tough question

[Mubariz Ahmed]

hey guys
i have this really hard maths (0580) question.....oct nov p43 q8 (e)
the answer is 9 but i cant get it.....plz help

 

[Potential]

In order to solve it you first have to rearrange h^-1 (x).
You start with replacing it with y= 3^x, the swapping the x and y around to make x = 3^y. Then you rearrange it with logs to get log(3)x = y
So log(3)x = 2 therefore x =3^2 = 9

 

[DarkEclipse]

In order to solve it you first have to rearrange h^-1 (x).
You start with replacing it with y= 3^x, the swapping the x and y around to make x = 3^y. Then you rearrange it with logs to get log(3)x = y
So log(3)x = 2 therefore x =3^2 = 9

We don't have logs in this syllabus.

So essentially, here's what you do.
h^-1(x) = 2.
This means that the inverse of h is 2.
Now, x=h(h^−1(x))=h(2) Means that the function of h is the inverse function of h.
So, now we can solve the equation!
h(2)= 3^2 =9!
No logarithms required!

 

[tosinwashere]

We don't have logs in this syllabus.

Do you think one would we be penalized if they used log? I sure hope not :/

 

[DarkEclipse]

Do you think one would we be penalized if they used log? I sure hope not :/

No, they don't
Don't worry