10 ^ x power is 3, 10 ^ y power is 4, find 10 ^ - Y / 2, find 10 ^ 3 (X-Y), find 10 ^ - x + 10 ^ - Y

10 ^ x power is 3, 10 ^ y power is 4, find 10 ^ - Y / 2, find 10 ^ 3 (X-Y), find 10 ^ - x + 10 ^ - Y

10^-x=10^x * 10^-1=3*(1/10)=3/10
10^-y=10^y * 10^-1=4*(1/10)=2/5
10^-x+10^-y=3/10+2/5=7/10
10^-y/2=10^-y * 10^1/2=2/5 * √10 =2√(2/5)

We know that the m power of a is equal to two and the nth power of a is equal to 8

a^(m+n)
=a^m*a^n
=2*8
=16

It is known that the m-th power of a is equal to 8, and the n-th power of a is equal to 32. Find the value of M + n power of A

M + n times of a = m times of a × n times of a, so it is equal to 256

If the even function f (x) and the odd function g (x) over R satisfy the power of F (x) + G (x) = e, then G (x)= The title is Hubei in 2011,

If the even function f (x) and the odd function g (x) over R satisfy the power of F (x) g (x) = e, then G (x)=
f(x) g(x)=e^x (1),
f(-x) g(-x)=e^(-x)
Because f (x) is an even function, G (x) is an odd function
f(x) -g(x)=e^(-x) (2)
(1) (2)
2g(x)=e^x-e^(-x)
g(x)=[e^x-e^(-x)]/2

Is the negative quartic function of y = x even or odd? What about minus cubic? What about the images?

F (x) = f (- x) is an even function, f (x) = - f (x) is an odd function, which can not be simply divided into two parts: one is odd function, the other is even function, and the other is neither odd function nor even function. It depends on whether the definition domain is consistent. For example, y = x, X is not equal to one and is a real number, which is even function which is neither odd function nor even function

Y = 3's - × th power - 3 / 1 - × power + 1 to judge whether it is an odd function or even function There should be a detailed process

Simplification
y=3^(-x)-1/3^(-x)
=3^(-x)-3^x
Let y = f (x)
f(x)=3^(-x)-3^x
f(-x)=3^(x)-3^(-x)=-[3^x-3^(-x)]=-f(x)
Because f (- x) = - f (x)
So it's an odd function

Is y = 1 + 2 / (x-1 of 2) odd, even, non odd or even

f(x)=1+2／（2^x-1）
f(-x) = 1 + 2/(2^(-x)-1)
= 1 - 2(2^x)/(2^x-1)
= 1 - 2( (2^x-1)/(2^x-1)+1/(2^x-1))
= -1 - 2/(2^x-1)
= -f(x)
F is an odd function

Find the definition domain of the following functions: (1) y = 2 to the (3-x) power; (2) y = 3 to the (2x + 1) power; (3) y = half to the power of 5x 1 / X important process of y = 0.7

Definition domain: x = R
Because x takes any value, questions 1, 2, 3 are meaningful
Y = 0.7 ^ (1 / x) domain x ≠ 0,

The known function f (x) = (1) 2x−1+1 2）x3． (1) Find the definition domain of F (x); (2) Judge the parity of F (x); (3) It is proved that f (x) > 0

(1) From the analytic expression of the function, we can get 2x-1 ≠ 0, and the solution is x ≠ 0. Therefore, the definition domain of function is {x | x ∈ R, and X ≠ 0}
(2) Obviously, the domain of the function is symmetric about the origin, f (- x) = (1)
2−x−1+1
2）（-x）3=（2x
1−2x+1
2）（-x）3 
=（2x−1+1
1−2x+1
2）（-x）3
=（-1+1
1−2x+1
2）（-x）3=-（1
2x−1+1
2）（-x）3=（1
2x−1+1
2）x3 =f（x），
Therefore, the function f (x) is even
(3) When x > 0, 1
2x−1+1
2＞1
2, X3 > 0, ν function f (x) = (1)
2x−1+1
2）x3 ＞0．
When x < 0, 1
2x−1＜-1，1
2x−1+1
2 ﹤ 0, x3 ﹤ 0, ﹤ function f (x) = (1
2x−1+1
2）x3 ＞0．
In conclusion, f (x) > 0

What is the definition domain of y = arcsin (LNX) + X / √ 2-x?

The domain of arcsin is [- 1,1], - 1