Question

Log7(2x+9)=log7(x^2+3x-18)

Answer 1

Решение на фото.........

Answer 2

To solve the equation, we need to eliminate the logarithms. Since both sides of the equation have the same base (log base 7), we can use the property that if log_b(a) = log_b(c), then a = c.

So, we have:

log7(2x + 9) = log7(x^2 + 3x - 18)

Now, using the property of logarithms, we can set the arguments of the logarithms equal to each other:

2x + 9 = x^2 + 3x - 18

Now, we have a quadratic equation:

x^2 + 3x - 18 - 2x - 9 = 0

Simplify:

x^2 + x - 27 = 0

Now, we can solve this quadratic equation by factoring or using the quadratic formula. Let's use the quadratic formula:

The quadratic formula is given by:

x = (-b ± √(b^2 - 4ac)) / 2a

where a = 1, b = 1, and c = -27.

Now, plug the values into the formula:

x = (-(1) ± √(1^2 - 4 * 1 * (-27))) / 2 * 1

x = (-1 ± √(1 + 108)) / 2

x = (-1 ± √109) / 2

So, we have two solutions:

1. x = (-1 + √109) / 2
2. x = (-1 - √109) / 2

These are the two solutions to the original equation.