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

[brendon2000]

very lost on hoe they derived some of the equations.

 

[Romsek]

it's a typo. They mean

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

 

[brendon2000]

thank you

 

[brendon2000]

ive been up with my friend to solve this qn

 

[Otis]

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.

Please read the forum's submission guidelines, too. Cheers!

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[HallsofIvy]

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 \(x\).