Question

Log7(х^2-х-7)=log7(-х+1)

Answer 1

$x^{2} -x-7=-x+1$
$x^{2} =8$ $x=2 \sqrt{2} x=-2 \sqrt{2}$

Answer 2

Log7(х^2-х-7)=log7(-х+1)
x²-x-7=-x+1   x²-x-7>0   -х+1>0
x²=8  x1=-2√2   x2=2√2
2√2+1>0  8+2√2-7>0
x2=2√2  -2√2+1√<0  не подходит
x=-2√2

Answer 3

To solve the equation log7(x^2 - x - 7) = log7(-x + 1), we need to use the properties of logarithms. The key property we'll use here is that if loga(b) = loga(c), then b = c.

So, in this case, we have:

log7(x^2 - x - 7) = log7(-x + 1)

Since the bases of both logarithms are the same (7 in this case), we can equate the arguments:

x^2 - x - 7 = -x + 1

Now, we have a quadratic equation to solve for x. To do that, we'll bring all terms to one side:

x^2 - x + x - 7 + 1 = 0

x^2 - 6 = 0

Now, factor the quadratic equation:

(x - 3)(x + 2) = 0

Now, set each factor to zero and solve for x:

x - 3 = 0 --> x = 3

x + 2 = 0 --> x = -2

So, the two solutions for the equation are x = 3 and x = -2. However, we need to check these solutions in the original equation to ensure they are valid, as logarithms are not defined for negative or zero arguments.

For x = 3: log7(3^2 - 3 - 7) = log7(9 - 3 - 7) = log7(-1)

The argument of the logarithm is negative (-1), which is not allowed, so x = 3 is not a valid solution.

For x = -2: log7((-2)^2 - (-2) - 7) = log7(4 + 2 - 7) = log7(-1)

Again, the argument of the logarithm is negative (-1), which is not allowed, so x = -2 is also not a valid solution.

Since both potential solutions are invalid, there are no real solutions to the original equation.