Answer:
The inverse for log₂(x) + 2 is - log₂x + 2.
Step-by-step explanation:
Given that
f(x) = log₂(x) + 2
Now to find the inverse of any function we put we replace x by 1/x.
f(x) = log₂(x) + 2
f(1/x) =g(x)= log₂(1/x) + 2
As we know that
log₂(a/b) = log₂a - log₂b
g(x) = log₂1 - log₂x + 2
We know that log₂1 = 0
g(x) = 0 - log₂x + 2
g(x) = - log₂x + 2
So the inverse for log₂(x) + 2 is - log₂x + 2.
solution is (x, y ) → (- 1, 4 )
y + 2x = 2 → (1)
y + 4x = 0 → (2)
subtract equation (1) from (2) to eliminate term in y
(y - y ) + (4x - 2x ) = (0 - 2 )
2x = - 2 ( divide both sides by 2 )
x = - 1
substitute x = - 1 in either (1) or (2) and solve for y
(1) → y - 2 = 2 ( add 2 to both sides )
y = 4
solution is (x, y ) → (- 1, 4 )
Answer:
-0.01
Step-by-step explanation:
48/5 or 9 3/5
12.8/1 x 3/4 = 38.4/4
Or multiply 12.8/1 by 10 ==> 128/10
128/10 x 3/4 = 384/40
Hope this helps!
