Question

Log2 (x^2-9)=log2 (4x+3)

Answer 1

Так как основания равны, то и

x^2-9=4x+3

x^2-9-4x-3=0

x^2-4x-12=0

D=b^2-4ac=16-4•1•(-12)=16+48=64

x1= (4-8)/2= -2 ( не подходит, так как x^2-9 > 0)

x2=(4+8)/2= 6

Ответ: 6

Answer 2

To solve this equation, we can use the fact that for any base "b" of logarithm, if logb(a) = logb(b), then a = b. So we can start by using this fact to rewrite the equation as:

x^2 - 9 = 4x + 3

Now we can simplify this equation by moving all the terms to one side:

x^2 - 4x - 12 = 0

Next, we can factor this quadratic equation to get:

(x - 6)(x + 2) = 0

Therefore, the solutions are x = 6 or x = -2. However, we need to check if these solutions are valid by plugging them back into the original equation and making sure that they don't result in any undefined values (i.e., negative or zero arguments inside the logarithm).

For x = 6: log2(6^2 - 9) = log2(4(6) + 3) log2(27) = log2(27) Both sides are equal, so x = 6 is a valid solution.

For x = -2: log2((-2)^2 - 9) = log2(4(-2) + 3) log2(5) = log2(-5) The left side is defined, but the right side is undefined since the logarithm of a negative number is not defined in the real numbers. Therefore, x = -2 is not a valid solution.

So the only solution to the equation is x = 6.