Question

Hi! does anyone know how to find the inverse function of y= log(base2) (x^2-3x-4)? I have been stuck on this question for hours!!!

Answer 1

are you sure it's log(base 2) ?

Answer 2

yeah i'm pretty sure

Answer 3

just do a flippity-flip-flip:

x = log(base2)(y^2 - 3y - 4)

so 2^x = y^2 - 3y - 4

so y^2 - 3y - 4 - 2^x = 0

use the quadratic formula:

y = (3 +- sqrt(17 + 4*2^x))/2

Answer 4

\[2^y = x^2 - 3x - 4\]\[x^2 - 3x - (4 + 2^y)=0\]\[x = \frac{1}{2}\left(3-\sqrt{2^{y+2} + 25}\right)\]

Answer 5

ergh...krebante is right...i was careless

Answer 6

Small typo: should be \[x = \frac{1}{2}\left(3 \pm \sqrt{2^{y + 2} + 25}\right).\]

Answer 7

thankyou so much! that was the answer on the maths book but I didnt know how it got there

Answer 8

wait, isn't it if x>4, f^-1=(3+((under square root)25+2^x+2)/2?

Answer 9

if you restrict x to x > 4, then you drop the minus sign, yes

Answer 10

but why is it to the power of x+2 instead of y+2

Answer 11

because the inverse function maps every point (x, y) to (y, x)

therefore, once you get x = 1/2(3 + sqrt(25 + 2^(y+2))), then you replace 'x' with 'y' and 'y' with 'x'

Answer 12

ok yes sorry about that stupid question haha