Answer:
3. see attached. Domain x > 1; Range all reals
4. √15 ≈ 3.87298
Explanation:
You want a graph and the domain and range of f(x) = 3log(2(x-1)) +4, and the value of (3x)^(x/10) when x satisfies 2^x -2^(x-1) = 16.
3. Log function
The function f(x) = 3log(2(x -1)) +4 represents horizontal and vertical scaling, as well as horizontal and vertical translation. The values in each case are different, so we can describe their effect by referencing their value.
1 — a horizontal right shift of the log function by 1 unit
2 — a horizontal compression of the log function by a factor of 2
3 — a vertical expansion of the log function by a factor of 3
4 — a vertical shift upward of the log function by 4 units.
Some points that you might graph for the parent log function would be (0.1, -1), (1, 0), (10, 1). The effect of these shifts is to transform them to ...
(x, y) ⇒ (x/2 +1, 3y +4)
(0.1, -1) ⇒ (1.05, 1)
(1, 0) ⇒ (1.5, 4)
(10, 1) ⇒ (6, 7)
The right shift 1 unit moves the domain to x > 1.
The range continues to be all real numbers.
4. (3x)^(x/10)
We can solve the given equation as follows:
2^x -2^(x -1) = 16
2^(x-1) · (2 -1) = 16
2^(x -1) = 2^4
x -1 = 4
x = 5
Then the expression you want the value of is ...
(3·5)^(5/10) = 15^(1/2) = √15 ≈ 3.87298
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