CA Foundation Exam > CA Foundation Questions > Log4(x2+x)-log4(x+1) = 2. Find xa)16b)0c)-1d)...
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Log4 (x2+x)-log4 (x+1) = 2. Find x
- a)16
- b)0
- c)-1
- d)none
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Log4(x2+x)-log4(x+1) = 2. Find xa)16b)0c)-1d)noneCorrect answer is opt...
Given equation: log4(x2 x)-log4(x+1) = 2
Step 1: Simplify the equation using logarithmic rules
log4(x2 x)-log4(x+1) = log4(x2 x / (x+1))
Step 2: Replace the logarithmic expression with its equivalent exponential form
log4(x2 x / (x+1)) = 2
4^2 = x2 x / (x+1)
16 = x2 x / (x+1)
Step 3: Solve for x
x2 x = 16(x+1)
x2 x = 16x + 16
x(x2 - 16) = -16
x(x-4)(x+4) = -16
Step 4: Check for extraneous roots
x = -4 and x = 4 are extraneous roots as they make the denominator of the logarithmic expression zero.
Hence, the only valid solution is x = -1.
Answer: Option (c) -1.
Step 1: Simplify the equation using logarithmic rules
log4(x2 x)-log4(x+1) = log4(x2 x / (x+1))
Step 2: Replace the logarithmic expression with its equivalent exponential form
log4(x2 x / (x+1)) = 2
4^2 = x2 x / (x+1)
16 = x2 x / (x+1)
Step 3: Solve for x
x2 x = 16(x+1)
x2 x = 16x + 16
x(x2 - 16) = -16
x(x-4)(x+4) = -16
Step 4: Check for extraneous roots
x = -4 and x = 4 are extraneous roots as they make the denominator of the logarithmic expression zero.
Hence, the only valid solution is x = -1.
Answer: Option (c) -1.
Community Answer
Log4(x2+x)-log4(x+1) = 2. Find xa)16b)0c)-1d)noneCorrect answer is opt...
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Log4(x2+x)-log4(x+1) = 2. Find xa)16b)0c)-1d)noneCorrect answer is option 'A'. Can you explain this answer?
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