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The equation log2 (3- x) + log2 (1- x) = 3 has
- a)One root
- b)Two root
- c)Infinite roots
- d)No root
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The equation log2 (3- x) + log2(1- x) = 3 hasa)One rootb)Two rootc)Inf...
log2 (3- x)(1- x) = 3 ⇒ (3- x)(1- x) = 8
⇒ x = 5 & x =-1,x = 5 fail ∴ x =-1
⇒ x = 5 & x =-1,x = 5 fail ∴ x =-1
Most Upvoted Answer
The equation log2 (3- x) + log2(1- x) = 3 hasa)One rootb)Two rootc)Inf...
Explanation:
Given Equation:
$log_2 (3 - x) + log_2 (1 - x) = 3$
Using Properties of Logarithms:
- $log_a (m) + log_a (n) = log_a (m*n)$
Applying the Property:
$log_2 [(3 - x)(1 - x)] = 3$
Expanding the Expression:
$(3 - x)(1 - x) = 2^3$
Solving the Equation:
$3 - 3x - x + x^2 = 8$
Rearranging Terms:
$x^2 - 4x - 5 = 0$
Using Quadratic Formula:
$x = \frac{4 \pm \sqrt{(-4)^2 - 4*1*(-5)}}{2*1}$
$x = \frac{4 \pm \sqrt{16 + 20}}{2}$
$x = \frac{4 \pm \sqrt{36}}{2}$
$x = \frac{4 \pm 6}{2}$
Therefore, the Roots are:
$x = \frac{4 + 6}{2} = 5$ (Valid)
$x = \frac{4 - 6}{2} = -1$ (Invalid as it makes the logarithm undefined)
Hence, the Equation has:
One root
Explanation:
Given Equation:
$log_2 (3 - x) + log_2 (1 - x) = 3$
Using Properties of Logarithms:
- $log_a (m) + log_a (n) = log_a (m*n)$
Applying the Property:
$log_2 [(3 - x)(1 - x)] = 3$
Expanding the Expression:
$(3 - x)(1 - x) = 2^3$
Solving the Equation:
$3 - 3x - x + x^2 = 8$
Rearranging Terms:
$x^2 - 4x - 5 = 0$
Using Quadratic Formula:
$x = \frac{4 \pm \sqrt{(-4)^2 - 4*1*(-5)}}{2*1}$
$x = \frac{4 \pm \sqrt{16 + 20}}{2}$
$x = \frac{4 \pm \sqrt{36}}{2}$
$x = \frac{4 \pm 6}{2}$
Therefore, the Roots are:
$x = \frac{4 + 6}{2} = 5$ (Valid)
$x = \frac{4 - 6}{2} = -1$ (Invalid as it makes the logarithm undefined)
Hence, the Equation has:
One root
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The equation log2 (3- x) + log2(1- x) = 3 hasa)One rootb)Two rootc)Infinite rootsd)No rootCorrect answer is option 'A'. Can you explain this answer?
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