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The number of real values of parameter k for which (log16x)2−log16x+log16k=0 will have exactly one solution is
- a)0
- b)2
- c)1
- d)4
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The number of real values of parameter k for which (log16x)2−log...
Understanding the Equation
The equation given is:
(log16x)² - log16x + log16k = 0.
Let y = log16x. Then, we can rewrite the equation as:
y² - y + log16k = 0.
This is a quadratic equation in terms of y.
Condition for Exactly One Solution
For a quadratic equation of the form ay² + by + c = 0 to have exactly one solution, the discriminant must be zero. The discriminant (D) is given by:
D = b² - 4ac.
In our equation:
- a = 1
- b = -1
- c = log16k
Thus, the discriminant becomes:
D = (-1)² - 4(1)(log16k) = 1 - 4log16k.
Setting the discriminant to zero for exactly one solution:
1 - 4log16k = 0.
This simplifies to:
log16k = 1/4.
Solving for k
To solve for k, we convert the logarithmic equation to its exponential form:
k = 16^(1/4).
Calculating this gives:
k = 2.
Finding Real Values of k
The value k = 2 is the only solution that satisfies the condition for exactly one solution of y.
Thus, there is only one real value of k that results in the quadratic equation having exactly one solution.
Final Answer
Therefore, the number of real values of parameter k for which the original equation has exactly one solution is 1. Thus, the correct option is C.
The equation given is:
(log16x)² - log16x + log16k = 0.
Let y = log16x. Then, we can rewrite the equation as:
y² - y + log16k = 0.
This is a quadratic equation in terms of y.
Condition for Exactly One Solution
For a quadratic equation of the form ay² + by + c = 0 to have exactly one solution, the discriminant must be zero. The discriminant (D) is given by:
D = b² - 4ac.
In our equation:
- a = 1
- b = -1
- c = log16k
Thus, the discriminant becomes:
D = (-1)² - 4(1)(log16k) = 1 - 4log16k.
Setting the discriminant to zero for exactly one solution:
1 - 4log16k = 0.
This simplifies to:
log16k = 1/4.
Solving for k
To solve for k, we convert the logarithmic equation to its exponential form:
k = 16^(1/4).
Calculating this gives:
k = 2.
Finding Real Values of k
The value k = 2 is the only solution that satisfies the condition for exactly one solution of y.
Thus, there is only one real value of k that results in the quadratic equation having exactly one solution.
Final Answer
Therefore, the number of real values of parameter k for which the original equation has exactly one solution is 1. Thus, the correct option is C.
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Community Answer
The number of real values of parameter k for which (log16x)2−log...
Let log16x=t
Hence, t2−t+log16k=0 and change k>0
The equation has exactly one solution. Hence, the discriminant must be zero.
Δ=1−4log16k
⇒0=1−4log16k
⇒log16k=1/4
⇒k=2
So, number of real values of k is 1 when k=2.
Hence, t2−t+log16k=0 and change k>0
The equation has exactly one solution. Hence, the discriminant must be zero.
Δ=1−4log16k
⇒0=1−4log16k
⇒log16k=1/4
⇒k=2
So, number of real values of k is 1 when k=2.
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