Find the domain of the function f (x) = log2 (a ^ X-2 ^ x * k) (a > = 2, and K is a constant)

Find the domain of the function f (x) = log2 (a ^ X-2 ^ x * k) (a > = 2, and K is a constant)

a^x-(2^x)*k>0
a^x>(2^x)*k
(1)a>2
When K0
xlga>xlg2+lgk
Domain x > LGK / (lga-lg2)
(2) a=2
When k = 1, a ^ x > (2 ^ x) * k does not hold, and the domain is an empty set,
It doesn't match the function definition. It's impossible

The domain of function y = f (x) is [- 1,1]. If K belongs to (0,1), then what is the domain of F (x) = f (x-k) + F (x + k)?
Please tell me more details. Thank you very much!

x-k∈[-1,1],x+k∈[-1,1]
Because K belongs to 0 to 1. That is, K is less than 1
So x1 ∈ (- 1 + K, 1 + k), X2 ∈ (- 1-k, 1-k)
among
-1 + k is greater than - 1-k, 1-k is less than 1 + K
So the domain is [- k-1, 1 + k]
I didn't see the answer. I found that D and a are the same. You may have the wrong number
Also, because the title says that K ∈ (0,1), so K cannot be equal to 1,
Then [- k-1, 1-k]

The domain of the function y = f (x) is [- 1,1]. If K belongs to (0,1), then the domain of F (x) = f (x-k) + F (x + k) is

The definition field of function y = f (x) is [- 1,1]
I.e. - 1

If K ∈ (0,1), then the domain of F (x) = f (x-k) + F (x + k) is ()
A [-k-1,1-k] B[k-1,k+1]
C[-k-1,k+1] D[-k-1,1-k]
The correct answer is a

X-k ∈ [- 1,1], x + K ∈ [- 1,1] because K belongs to 0 to 1. That is, K is less than 1, so x1 ∈ (- 1 + K, 1 + k), X2 ∈ (- 1-k, 1-k) where - 1 + k is greater than - 1-k, 1-k is less than 1 + K, so the domain of definition is [- k-1,1 + k]. I don't see the answer, I find that D and a are the same, you may have the wrong number, because the title says

2 (ab2-2a2b) - 3 (ab2-a2b) + (2ab2-2a2b) where a = 2, B = 1

The original formula = 2ab2-4a2b-3ab2 + 3a2b + 2ab2-2a2b = ab2-3a2b, ∵ a = 2, B = 1. The original formula = 2 × 1-3 × 4 × 1 = - 10

The total current of the circuit is 5A when 4 and 6 ohm resistors are connected in parallel to a power supply. If they are connected in series to the same power supply, what is the current in the circuit?
What are the voltages at both ends of the two resistors?

Circuit voltage = current * resistance
Resistance = 1 / [(1 / 4) + (1 / 6)] 12 / 5
Therefore, the voltage = 5 * (12 / 5) = 12 v
After series connection, resistance = 4 + 6 = 10 ohm
Current = voltage / resistance = 12 / 10 = 1.2 A
The voltages at both ends of the resistor are 1.2 * 4 = 4.8V and 1.2 * 6 = 7.2V respectively

If the slide AC is placed horizontally, it is just as long as ab. given the height of the slide CE = 3cm, CD = 1m, calculate the length of the slide AC

Let the length of AC be x m, ∵ AC = AB, ∵ AB = AC = x m, ∵ EB = CD = 1 m, ∵ AE = (x-1) M. in RT △ ace, ac2 = CE2 + AE2, that is, X2 = 32 + (x-1) 2, the solution is: x = 5,