# i dont know how they derive the answer at "hence" can someone explain it to me

#### brendon2000

##### New member
very lost on hoe they derived some of the equations.

#### Attachments

• 160.5 KB Views: 33
• 107.5 KB Views: 30

#### Romsek

##### Full Member
it's a typo. They mean

$$\displaystyle f^{-1}(x) = 2^{x+3}$$

thank you

#### brendon2000

##### New member
ive been up with my friend to solve this qn

#### Otis

##### Senior Member
ive been up with my friend to solve this qn
You're talking about question B3 part (b), yes?

Can you post the work that you and your friend have done so far? We would like to see where you're stuck.

Problem B3 (b) says "Given $$f(x)= log_2(x)- 3$$ solve the following for x: $$f^{-1}(x-1)= 4^x+3$$." Actually it isn't necessary to find $$f^{-1}$$. Taking $$f$$ of both sides gives $$f(f^{-1}(x- 1))= x- 1= log_2(4^x+ 3)$$.
We can write this as $$2^{x- 1}= 2^x/2= 4^x+ 3= 2^{2x}+ 3$$. Let $$y= 2^x$$ so this becomes $$y/2= y^2+ 3$$ or $$2y^2- y+ 6= 0$$. Solve that using the quadratic equation then solve $$2^x= y$$ for $$x$$.