Question

Log8 x+ log8 (x+2)=1

Answer 1

Odz}x+2>0}x>-2
x^2+2x-8=0
d=4+32=36
x1=(-2+6)/2=2
x2=(-2-6)/2=-4-не входит в одз
otvet:2

Answer 2

To solve the equation log8(x) + log8(x + 2) = 1, we'll use logarithm rules to simplify the expression and then solve for x.

Step 1: Combine the logarithms using the product rule: log_b(a) + log_b(c) = log_b(ac) log8(x) + log8(x + 2) = log8(x * (x + 2))

Step 2: Apply the logarithm rule log_b(b^y) = y to get rid of the logarithm on the right side: log8(x * (x + 2)) = 1

Step 3: Rewrite the equation using the definition of logarithms: b^y = x is equivalent to log_b(x) = y 8^1 = x * (x + 2)

Step 4: Simplify the equation: 8 = x^2 + 2x

Step 5: Rearrange the equation in standard quadratic form (ax^2 + bx + c = 0): x^2 + 2x - 8 = 0

Step 6: Solve the quadratic equation for x using factoring or the quadratic formula: The quadratic equation factors to: (x + 4)(x - 2) = 0

Set each factor to zero and solve for x: x + 4 = 0 or x - 2 = 0

For x + 4 = 0, subtract 4 from both sides: x = -4

For x - 2 = 0, add 2 to both sides: x = 2

Step 7: Check the solutions: Substitute each value back into the original equation to ensure they are valid solutions.

For x = -4: log8(-4) + log8(-4 + 2) = log8(-4) + log8(-2) The logarithm of a negative number is not defined for real numbers, so x = -4 is not a valid solution.

For x = 2: log8(2) + log8(2 + 2) = log8(2) + log8(4) The logarithm of a positive number is well-defined, and the equation holds true: 1 + 1 = 1 (True)

Therefore, the only valid solution to the equation is x = 2.