Question

Lq(x-1)+ lq(x-2,5)=1

Answer 1

Lg(x-1)(x-2,5)=lg10. Одз:х>1 и х>2,5
X^2-3,5x+2,5-10=0
X^2-3,5x-7,5=0 |*10
10x^2-35x-75=0
D=35^2+4*75*10=25*49+25*12*10=25*(49+120)=25*169
X1,2=(35+-65)/20
X1=5
X2=-1 -не подходит
Ответ :х=5

Answer 2

The equation you provided is:

Lq(x-1) + Lq(x-2.5) = 1

where Lq is the natural logarithm function.

To solve this equation for x, we can use algebraic manipulation to isolate x on one side of the equation.

First, we can combine the two logarithms on the left-hand side using the product rule of logarithms:

Lq[(x-1)(x-2.5)] = 1

Next, we can exponentiate both sides using the definition of logarithms:

e^1 = (x-1)(x-2.5)

Simplifying the right-hand side, we get:

e = x^2 - 3.5x + 2.5

Rearranging the terms, we get:

x^2 - 3.5x + (2.5 - e) = 0

Now we can solve this quadratic equation using the quadratic formula:

x = [3.5 ± sqrt(3.5^2 - 4(1)(2.5 - e))]/(2(1))

Simplifying further, we get:

x = [3.5 ± sqrt(12.25 + 4e - 10)]/2

x = [3.5 ± sqrt(4e + 2.25)]/2

Therefore, the solutions to the equation Lq(x-1) + Lq(x-2.5) = 1 are:

x = [3.5 + sqrt(4e + 2.25)]/2 or x = [3.5 - sqrt(4e + 2.25)]/2

Note that we can only take the natural logarithm of positive real numbers, so the expression inside each logarithm must be greater than zero. We can check that both solutions satisfy this condition.