Question

Log2(2x+1)=log2(9x+17)-log2(x+5)

Answer 1

Ответ:

Ураааааапа 3546+6×54₸-•==%346

Answer 2

To solve the given equation:

log2(2x + 1) = log2(9x + 17) - log2(x + 5)

We can simplify it using logarithmic properties. The subtraction of logarithms can be expressed as the division of their arguments, so we have:

log2(2x + 1) = log2((9x + 17) / (x + 5))

Since the base of the logarithm is the same on both sides (base 2), we can remove the logarithm and equate the arguments:

2x + 1 = (9x + 17) / (x + 5)

Now we can solve for x. To do that, we'll begin by multiplying both sides by (x + 5) to eliminate the fraction:

(x + 5)(2x + 1) = 9x + 17

Expanding the left side:

2x^2 + 11x + 5 = 9x + 17

Rearranging the equation:

2x^2 + 11x - 9x + 5 - 17 = 0

2x^2 + 2x - 12 = 0

Dividing the equation by 2 to simplify:

x^2 + x - 6 = 0

Now we can factor the quadratic equation:

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

Setting each factor equal to zero:

x + 3 = 0 -> x = -3

x - 2 = 0 -> x = 2

Therefore, the equation log2(2x + 1) = log2(9x + 17) - log2(x + 5) has two solutions: x = -3 and x = 2.