Question

Lg(x+1.5) = - lg xЧто вообще сделать тут? Я попробовала (x+1.5)= - х, но ответ какой-то стремный получился.(

Answer 1

Lg(x+1.5) = - lg x, lg(x+1.5) =  lg 1/x, x+1.5 =  1/x, x^2+1.5x-1 =  0,  x1=-2, x2=0,5
ответ: x=0,5

Answer 2

lg(x+1.5) = - lg x
ОДЗ x>-1.5
 x>0
lg(x+1.5) = lg x^-1
lg (x+1.5) = lg 1/x
x+3/2=1/x приводиь к общему знаменателю
2x^2+3x-2=0
x12=(-3+-квадрат(9+4*2*2))/4= (-3+-5)/4=-2  1/2
x=-2 не подходит по ОДЗ x>0
x=1/2

Answer 3

Solving the Equation lg(x+1.5) = -lg(x)

To solve the equation lg(x+1.5) = -lg(x), we can use logarithmic properties to simplify and solve for x.

First, let's rewrite the equation using the properties of logarithms. The equation lg(x+1.5) = -lg(x) can be rewritten as:

lg(x+1.5) + lg(x) = 0

Now, we can use the property that states: lg(a) + lg(b) = lg(a * b). Applying this property to the equation, we get:

lg(x * (x+1.5)) = 0

Since lg(1) = 0, the only way for the logarithm to equal 0 is if the argument of the logarithm equals 1. Therefore, we have:

x * (x+1.5) = 1

Now, we can solve for x by expanding and rearranging the equation:

x^2 + 1.5x - 1 = 0

This is a quadratic equation, and we can solve for x using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Where a = 1, b = 1.5, and c = -1.

After solving for x, we can check the solutions to ensure they satisfy the original equation.

Let's calculate the solutions for x using the quadratic formula.