Algebra - Logarithms - Previous Year CAT/MBA Questions
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Consider the equation , where x is a real number log5(x - 2) = 2log25(2x - 4).
For how many different values of x does the given equation hold?
- (a)
0
- (b)
1
- (c)
2
- (d)
4
- (e)
Infinitely many
Answer: Option A
Text Explanation :
Given, log5(x - 2) = 2log25(2x - 4).
⇒ log5(x - 2) = 2log52(2x - 4).
⇒ log5(x - 2) = 2/2 × log5(2x - 4).
⇒ log5(x - 2) = log5(2x - 4).
⇒ x - 2 = 2x - 4
⇒ x = 2
Now, for x = 2, log5(x - 2) will not be defined. Hence, x = 2 cannot be the solution
∴ We get no solution for the given equation.
Hence, option (a).
Workspace:
If log4 m + log4 n = log2 (m + n) where m and n are positive real numbers, then which of the following must be true?
- (a)
+ = 1
- (b)
m = n
- (c)
m2 + n2 = 1
- (d)
+ = 2
- (e)
No values of m and n can satisfy the given equation
Answer: Option E
Text Explanation :
log4 mn = log2 (m + n)
= (m + n)
Squaring on both sides
m2 + n2 + mn = 0
Since m, n are positive real numbers, no value of m and n satisfy the above equations.
Workspace:
equals which of the following?
- (a)
-5
- (b)
-3
- (c)
None of the others
- (d)
-4
- (e)
-2
Workspace:
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