Question

Log3(1-6x)=log3(17-x2)

Answer 1

$log_{3}(1 - 6x) = log_{3}(17 - {x}^{2} ) \\ 1 - 6x = 17 - {x}^{2} \\ {x }^{2} - 6x - 16 = 0 \\ (x - 8)(x + 2) = 0 \\ x1 = 8 \\ x2 = - 2$
ОДЗ
$\\ \left \{ {{1 - 6x > 0} \atop {17 - x ^{2} > 0}} \right.$
ОДЗ удовлетворяет лишь
$x = - 2$
Ответ
$x = - 2$

Answer 2

To solve the given equation, we can start by using the fact that if log base b of x equals log base b of y, then x equals y. In other words, we can set the expressions inside the logarithms equal to each other:

log3(1-6x) = log3(17-x^2)

1-6x = 17 - x^2 (using the fact that log base 3 of a number equals x if and only if 3^x equals that number)

Rearranging this equation, we get:

x^2 - 6x - 16 = 0

We can solve this quadratic equation using the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

In this case, a = 1, b = -6, and c = -16. Substituting these values into the quadratic formula, we get:

x = (6 ± sqrt(6^2 - 4(1)(-16))) / 2(1)

Simplifying:

x = (6 ± sqrt(100)) / 2

x = (6 ± 10) / 2

x = 8 or x = -2

However, we need to check whether both of these solutions satisfy the original equation or not:

For x = 8:

log3(1-6x) = log3(1-6(8)) = log3(-47) is not defined, so x = 8 is not a valid solution.

For x = -2:

log3(1-6x) = log3(1-6(-2)) = log3(13)

log3(17-x^2) = log3(17-(-2)^2) = log3(13)

Since both expressions evaluate to log3(13), the solution x = -2 is valid.

Therefore, the only solution to the given equation is x = -2.