Question

Lg'2(x+2)-3lg(x+2)+2=0​

Answer 1

Ответ:

Объяснение:

x>-2

lg(x+2)=t,

t²-3t+2=0

D=9-8=1

t1= (3-1)/2= 1

t2= (3+1)/2= 2

1) lg(x+2)=1

x+2=10

x1=8

2) lg(x+2)=2

x+2= 100

x2= 98

Answer 2

To solve the equation lg(2(x+2)) - 3lg(x+2) + 2 = 0, we can use logarithmic properties and algebraic manipulations. Here's the step-by-step solution:

Step 1: Combine the logarithmic terms using the rules of logarithms. lg(2(x+2)) - lg((x+2)^3) + 2 = 0

Step 2: Apply the quotient rule of logarithms to simplify further. lg((2(x+2))/(x+2)^3) + 2 = 0

Step 3: Simplify the expression inside the logarithm. lg(2/(x+2)^2) + 2 = 0

Step 4: Convert the equation into exponential form. 10^(lg(2/(x+2)^2)) = 10^(-2)

Step 5: Simplify the left side by canceling out the logarithm and exponential. 2/(x+2)^2 = 0.01

Step 6: Multiply both sides by (x+2)^2 to eliminate the denominator. 2 = 0.01(x+2)^2

Step 7: Expand and simplify the equation. 2 = 0.01(x^2 + 4x + 4) 2 = 0.01x^2 + 0.04x + 0.04

Step 8: Move all terms to one side to form a quadratic equation. 0.01x^2 + 0.04x + 0.04 - 2 = 0.01x^2 + 0.04x - 1.96 = 0

Step 9: Multiply the entire equation by 100 to eliminate the decimal. x^2 + 4x - 196 = 0

Step 10: Solve the quadratic equation using factoring, completing the square, or the quadratic formula. In this case, we will use the quadratic formula. x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 1, b = 4, and c = -196. x = (-4 ± √(4^2 - 4(1)(-196))) / (2(1)) x = (-4 ± √(16 + 784)) / 2 x = (-4 ± √800) / 2 x = (-4 ± 20√2) / 2 x = -2 ± 10√2

Therefore, the solutions to the equation lg(2(x+2)) - 3lg(x+2) + 2 = 0 are: x = -2 + 10√2 and x = -2 - 10√2.