The intersection coordinates of positive scale function y = 3x and inverse scale function y = 1 / X

The intersection coordinates of positive scale function y = 3x and inverse scale function y = 1 / X

There are two intersections:
A (one third of the root, three root)
B (negative root one third, negative root three)
Solve the equation a ^ (lgx). X ^ (LGA) - 2 [a ^ (lgx) + x ^ (LGA)] + 3 = 0
Let a ^ (lgx) = x ^ (LGA) take the logarithm with a as the base on both sides of a ^ (lgx) = x ^ (LGA) have lgx = loga (x ^ (LGA)) = LGA * logax = LGA * lgx / LGA = lgx, let a ^ (lgx) =. X ^ (LGA) = a, then the original formula is a * A-4 * a + 3 = 0, (A-3) (A-1) = 0A = 1, x ^ (LGA) = 1, take the logarithm with 10 as the base on both sides, LGA * LG
How many intersections are there between the positive scale function y = 3x and the inverse scale function y = 2 / x?
If y = 3x and y = 2 / X are used to calculate the intersection coordinates, we can get: 3x = 2 / x3x & # 178; = 2x & # 178; = 2 / 3x = ± √ (2 / 3) x = ± (√ 6) / 3, when x = (√ 6) / 3, y = √ 6; when x = - (√ 6) / 3, y = - √ 6, so the positive proportion function y = 3x and y = 2 / X have two intersections, which are [(√ 6) / 3, √ 6] and [- (√ 6) / 3, √ 6]
Solution equation (3 + lgx) ^ (1-lgx) = 1
(3+lgx)^(1-lgx)=1
(1) 3+lgx=1
Then lgx = - 2
∴ x=1/100
(2) 1-lgx = 0 and 3 + lgx ≠ 0
∴ lgx=1
That is, x = 10
To sum up, the solution of the equation is x = 1 / 100 or x = 10
How many intersections are there between the inverse scale function y = 2 / X and the positive scale function y = - 3x?
There is no intersection between two function images
analytic method
Solving equation 2 / x = - 3x
x²=-2/x
There is no intersection, the inverse scale function is in quadrant 1,4, and the positive scale function is in quadrant 2,3
2/x=-3x
unsolvable
So there's no intersection
Analysis: to calculate the intersection point is to calculate the common point of two functions, that is, Y1 = Y2, X1 = X2, that is, 2 / x = - 3x
Solve the equation lgx = 2lg5
lgx=2lg5=lg5^2=lg25
x=25
x=25
x=25
lgx=2lg5
lgx=lg25
x =25
lgx=lg（5*5）；x=25
2lg5=lg25
Then lgx = 2lg5, we can get x = 25
It's twenty-five
^What is the power of the sign
1. (1 / 2) ^ x times 8 ^ 2x = 4
2. 5 ^ 2x-6 times 5 ^ x + 5 = 0
3. 3^x-3^-x=80/9
4. 2logx25-3log25x = 1 (2 times x is the logarithm of base 25 - 3 times 25 is the logarithm of base x = 1)
5. Log7 (log3x) = - 1 (log multiply by 7 (logarithm of x = - 1)
If 2lg5 = LG25, then x = 25
As shown in the figure, it is known that the image of positive scale function y = x and inverse scale function y = 1x intersects at two points a and B. (1) the coordinates of two points a and B are obtained; (2) according to the image, the range where the value of positive scale function is larger than that of inverse scale function x is obtained
(1) According to the meaning of the problem, the coordinates of a and B satisfy the solution of the equation system y = xy = LX, X1 = 1y1 = 1, X2 = − 1Y2 = − 1, the coordinates of a and B are a (1,1), B (- 1, - 1); (2) according to the image, when - 1 < x < 0 or x > 1, the value of positive proportion function is greater than that of inverse proportion function
The solution set of the equation x ^ (lgx) = 5 * 2 ^ (lgx ^ (2) - 1) is
Take the logarithm of 10 as the base on both sides of the equation,
(lgx)^2 = lg5 + (2lgx -1)lg2
(lgx)^2 - 2lg2lgx +lg(2/5) = 0
(lgx)^2 - 2lg2lgx +(2lg2-1) =0
[lgx -(2lg2-1)](lgx -1) =0
therefore
Lgx = 2lg2-1 or lgx = 1
X = 2 / 5 or x = 10
The solution of the original equation is {10,2 / 5}
Take the logarithm of 10 as the base on both sides of the equation,
(lgx) ^ 2 = lg5 + (2lgx - 1) LG2
Another question is how to transform LG (2 / 5) to (2lg2-1)?
Taking the logarithm of 10 as the base on both sides of the equation, we get lgx ^ (lgx) = LG (5 * 2 ^ (lgx ^ 2-1)) because lgx ^ a = a * lgx (lgx) * (lgx) = lg5 + LG2 ^ (lgx ^ 2-1) (lgx) ^ 2 = lg5 + (2lgx - 1) LG2 and LG (2 / 5) = LG ((2 * 2) / (5 * 2)) = LG (2 * 2) - LG (5 * 2) = 2lg2-1
lgx^(lgx)=lg(5*2^(lgx^2-1))
lgx^A=A*lgx
(lgx)*(lgx)=lg5+lg2^(lgx^2-1)
(lgx)^2 = lg5 + (2lgx -1)lg2
Given that the positive scale function y = ax and the inverse scale function y = BX have no intersection point in the same coordinate system, then the relationship between a and B is______ .
If the positive scale function y = ax and the inverse scale function y = BX have no intersection in the same coordinate system, then the relationship between a and B is different
Given a = {x | y = LG (4-x2)}, B = {y | y ＞ 1}, then a ∩ B = ()
A. {x | - 2 ≤ x ≤ 1} B. {x | - 1 ＜ x ＜ 2} C. {x | - 2} D. {x | - 2 ＜ x ＜ 1 or X ＞ 2}
∫ set a = {x | y = LG (4-x2)} = {x | 4-x2 ＞ 0} = {x | - 2 ＜ x ＜ 2}, B = {y | y ＞ 1}, a ∩ B = {x | 1 ＜ x ＜ 2}