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Exponential problem in compulsory math course 1 for senior one.
What is the domain of 1 power of the function f (x) = (half 1)? What is the scope? The domain is x≠0, and as for y>0.

2. Equation: the solution of 2x+ 1 to the power of a = 1 4. It can only be expressed by logarithm, ㏒a 1/4=2x+ 1, because there are two unknowns in a formula, and there are no other restrictions, so it is impossible to have a solution.

3. the square of x of function y =a-3x+ 1 (0

4. Inequality: 2x-65438+2-2 to the 0th power >; 0 to find the range of X. It is deduced from the original formula that the quadratic power of 2x- 1 >: 2, 2 = 2 1 power, and the 2x- 1 power of 2 is increasing function, so the value range of X is 2x-1>; 1,x & gt 1

5. Equation: the solution of X-2 power x+2 -32 =0 of 4. Derived from the original formula that 2x power of 2-2x+2x power of 2-5 =0. There are only three numbers, the base is 2, and the sum is 0, so the exponents of two terms with equal symbols are equal, and the exponent of an item with a single symbol is any one of the other two+1 (this is the characteristic of base 2, you can try it yourself, this is beyond the higher range), so x+2=5, and x=3.

6. the function y = a+ 1 to the power of x-2 (a >;); O) Find which vertex the function passes through? You can only take a special point, that is, the point that crosses (0,2) when x=2.