Question

Log (2x^2-5x+3) по основании 6x^2-x-1=0

Answer 1

Ели логарифм равен 0, то 2x^2-5x+3=0
x^2-2*5x/4+25/16+24/16-25/16=0(х-5/16)^2=1/16
Два решения х=5/16-4/16=1/16
и х=5/16+4/16=9/16
Проверяем не станет ли основние логарифма отрицательным или нулевым.
6/256-1/16-1=6/256-16/256-1  меньше 0. Значит первое решение не годится.
486/256-144/256-256/256=86/256 больше 0.
Ответ: х=9/16

Answer 2

To find the value of log(2x^2 - 5x + 3) with base 6x^2 - x - 1 = 0, we need to solve the equation 6x^2 - x - 1 = 0 first and then substitute the solution into the logarithmic expression.

Let's solve the equation 6x^2 - x - 1 = 0 using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

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

x = (-(-1) ± √((-1)^2 - 4 * 6 * (-1))) / (2 * 6) = (1 ± √(1 + 24)) / 12 = (1 ± √25) / 12 = (1 ± 5) / 12

We have two possible solutions:

1. x = (1 + 5) / 12 = 6/12 = 1/2
2. x = (1 - 5) / 12 = -4/12 = -1/3

Now, let's substitute each solution into the logarithmic expression log(2x^2 - 5x + 3) with base 6x^2 - x - 1 = 0:

1. For x = 1/2: log(2(1/2)^2 - 5(1/2) + 3) with base 6(1/2)^2 - (1/2) - 1

Simplifying the expression: log(1/2 - 5/2 + 3) with base 3/2 - 1/2 - 1 log(-1) with base 1 The logarithm of a negative number is undefined in the real number system. Therefore, the logarithm is undefined for this solution.

2. For x = -1/3: log(2(-1/3)^2 - 5(-1/3) + 3) with base 6(-1/3)^2 - (-1/3) - 1

Simplifying the expression: log(2(1/9) + 5/3 + 3) with base 6(1/9) + 1/3 - 1 log(2/9 + 5/3 + 3) with base 2/3 + 1/3 - 1 log(2/9 + 15/9 + 27/9) with base 2/3 log(44/9) with base 2/3

Thus, log(2x^2 - 5x + 3) with base 6x^2 - x - 1 = 0 is undefined for x = 1/2 and log(44/9) with base 2/3 for x = -1/3.