Question

Log6(x-2)+log6(x-1)=1 очеееень срочно

Answer 1

Вот держи, вроде так

Answer 2

Log6(x-2)+log6(x-1)=1, x принадлежит (2, + бесконечность)
log6((x-2)*(x-1))=1
log6(x^2-x-2x+2)=1
log6(x^2-3x+2)=1
x^2-3x+2=6^1
x^2-3x+2-6=0
x^2-3x-4=0
Находим D: x1=4  и x2= -1
x2= -1 не подходит по условию 
Ответ: x=4

Answer 3

Solving the Equation Logarithmic Equation

To solve the equation Log6(x-2) + log6(x-1) = 1:

We can use the properties of logarithms to simplify the equation and solve for x.

First, we can combine the logarithms using the product rule of logarithms, which states that log_b(M) + log_b(N) = log_b(M * N).

So, the given equation can be rewritten as: log6((x-2)(x-1)) = 1

Next, we can rewrite the equation in exponential form. The logarithmic equation log_b(M) = N is equivalent to b^N = M.

Therefore, the equation becomes: 6^1 = (x-2)(x-1)

Solving for x: 6 = (x-2)(x-1) 6 = x^2 - 3x + 2

This is a quadratic equation, and we can solve for x by setting the equation equal to zero and factoring or using the quadratic formula.

Let's solve for x using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)

Where a = 1, b = -3, and c = 2.

Plugging these values into the quadratic formula: x = (-(-3) ± √((-3)^2 - 4(1)(2))) / (2*1) x = (3 ± √(9 - 8)) / 2 x = (3 ± √1) / 2 x = (3 ± 1) / 2

So, we have two possible solutions for x: x = (3 + 1) / 2 = 4 / 2 = 2 x = (3 - 1) / 2 = 2 / 2 = 1

Therefore, the solutions to the equation Log6(x-2) + log6(x-1) = 1 are x = 2 and x = 1.

This is the detailed solution to the given logarithmic equation. If you have any further questions or need additional assistance, feel free to ask!