Find equation 2 (log3 ^ x) ^ 2 + log3 ^ x-3 = 0

Find equation 2 (log3 ^ x) ^ 2 + log3 ^ x-3 = 0

2(log3^x)^2+log3^x-3=0
Let y = log3 ^ X
2y^2+y-3=0
(2y+3)(y-1)=0
y=3/2,y=1
log3^x=3/2
3^x=10^(3/2)
x=log10^(3/2)/log3
log3^x=1
3^x=10
x=log10/log3

Solve the equation: 6.8-1.2x = 6.542 + 3x = 69 6.5 × 8 + 5x = 79 1, the difference between 5 times of Y and 2 times of Y is 2.4, find y

—0.25 9 5.4 y=0.8

6 × (Y-8) = 1.2
Add two more questions, 0.7 times (4-x) = 1.758 divided by 2x = 1.6 to solve the equation

0.6×(y-8)=1.2
y-8=1.2÷0.6
y-8=2
y=10
Hope to adopt

The solution equation is 0.75: (x + 3) = 2:8

0.75:(x+3)=2:8
0.75*8=2(x+3)
6=2(x+3)
3=x+3
x=0

Solving equation 1 / 9 = 0.8: X

1/9=0.8:x
1/9=4/5x
x=5/36

Where log3 (x + 3) = the root of 3x?

The deformation is 3 ^ 3x = x + 3, that is 27 ^ x = x + 3
Draw the image, there are two intersections, there are two real roots

The equation log3 ^ x = - 3x has two positive roots, one positive root, one negative root, two negative roots and only one real root

A negative root

Given that the domain of function f (X & # 178;) is [1,2], find the domain of function f (x)

Given that the domain of function f (X & # 178;) is [1,2], find the domain of function f (x)
x∈ [1,2]
x²∈ [1,4]
The definition field of function f (x) is [1,4]

Given that the domain of definition of function f (X & # 178; - 2) is [1, + ∞), find the domain of definition of function f (x / 2)
The domain of definition of function f (X & # 178; - 2) is x ∈ [1, + ∞),
-1 ≤ x²－2
Why?
The domain of X & # 178; - 2 is not [1, + ∞)?:

The domain is the scope of X
So x > = 1
Then x & # 178; > = 1
So x & # 178; - 2 > = - 1

F (x) domain R. satisfy f (x + 2) (1-f (x)) = 1 + F (x). Prove: F (x) is a periodic function

f(x+2)(1-f(x))=1+f(x).
f(x+2)=[1+f(x)]/[1-f(x)] a
f(x)=[1+f(x-2)]/[1-f(x-2)] b
Substituting B into a, we can get f (x + 2) = - 1 / F (X-2) C
Similarly, f (X-2) = - 1 / F (X-6) d
Substituting d into C, f (x + 2) = f (X-6)
So f (x) = f (X-8)
The minimum period is 8