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The possible values of x for the equation log2 x2 + logx 2 = 3 are
- a)2 and √2
- b)1 and √2
- c)1 and 2
- d)2 and 4
- e)√2 and 4
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
The possible values of x for the equation log2 x2 + logx 2 = 3 area)2 ...
First, we simplify the equation using the laws of logarithms:
log2 x2 + log2 (1/log2 x) = 3
log2 (x2 * 1/log2 x) = 3
log2 (x * log2 x) = 3
x * log2 x = 2^3
x * log2 x = 8
We can then use trial and error or a graphing calculator to find the solutions to this equation. One solution is x = 2, since 2 * log2 2 = 2. Another solution is x ≈ 43.56, which can be found using a graphing calculator or by using iterative methods to approximate the solution.
log2 x2 + log2 (1/log2 x) = 3
log2 (x2 * 1/log2 x) = 3
log2 (x * log2 x) = 3
x * log2 x = 2^3
x * log2 x = 8
We can then use trial and error or a graphing calculator to find the solutions to this equation. One solution is x = 2, since 2 * log2 2 = 2. Another solution is x ≈ 43.56, which can be found using a graphing calculator or by using iterative methods to approximate the solution.
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The possible values of x for the equation log2 x2 + logx 2 = 3 area)2 ...
log2 x2 + logx 2 = 3
⇒
+ logx 2 = 3
⇒ 2 + (logx 2)2 - 3 logx 2 = 0
Put y = logx 2
⇒ y2 - 3y + 2 = 0
⇒ y = 1, 2
For y = 1, logx 2 = 1 or x = 2
For y = 2, logx 2 = 2 or x = √2
Hence, there are two valid values of x, i.e. x = 2, √2.
⇒
+ logx 2 = 3⇒ 2 + (logx 2)2 - 3 logx 2 = 0
Put y = logx 2
⇒ y2 - 3y + 2 = 0
⇒ y = 1, 2
For y = 1, logx 2 = 1 or x = 2
For y = 2, logx 2 = 2 or x = √2
Hence, there are two valid values of x, i.e. x = 2, √2.
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The possible values of x for the equation log2 x2 + logx 2 = 3 area)2 and√2b)1 and √2c)1 and 2d)2 and 4e)√2and 4Correct answer is option 'A'. Can you explain this answer?
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