Question

Log_2(x^2-3x+10)=3

Answer 1

Х2-3Х+10=8,
Х2-3Х+2=0,
по теореме виета Х=2 и Х=1

Answer 2

log_2(x^2-3x+10)=3
x^2-3x+10=2^3
x^2-3x+10=8
X^2-3x+10-8=0
x^2-3x+2=0
D=9-4*2*1=9-8=1
x1=3-1/2=1
x2=3+1/2=2

Answer 3

To solve the equation log_2(x^2 - 3x + 10) = 3, we can rewrite it in exponential form. In logarithmic equations, the base of the logarithm becomes the base of the exponential expression, and the logarithmic equation can be rewritten as:

2^3 = x^2 - 3x + 10

Simplifying further:

8 = x^2 - 3x + 10

Rearranging the equation:

x^2 - 3x + 10 - 8 = 0

x^2 - 3x + 2 = 0

To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring may not be straightforward, so let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For the equation x^2 - 3x + 2 = 0, we have:

a = 1, b = -3, and c = 2

Substituting these values into the quadratic formula:

x = (-(-3) ± √((-3)^2 - 4(1)(2))) / (2(1))

Simplifying:

x = (3 ± √(9 - 8)) / 2

x = (3 ± √1) / 2

x = (3 ± 1) / 2

x1 = (3 + 1) / 2 = 4 / 2 = 2

x2 = (3 - 1) / 2 = 2 / 2 = 1

Therefore, the solutions to the equation log_2(x^2 - 3x + 10) = 3 are x = 2 and x = 1.