solve qno 13 and 14 please
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Answer:
2 & 81
Step-by-step explanation:
Q 12
1 raise to any power is 1
=> x - 1 = 1
=> x = 2
Equating power as base is same
but x - 1 > 0 as mod is on LHS
log₃x² - 2logₓ9 = 7
=> 2log₃x - 2logₓ3² = 7
=> 2log₃x - 4logₓ3 = 7
Let say log₃x = y => logₓ3 = 1/y
=> 2y - 4/y = 7
=> 2y² - 4 = 7y
=> 2y² - 7y - 4 = 0
=> 2y² - 8y + y - 4 = 0
=> 2y(y -4) + 1(y - 4) = 0
=> (2y + 1)(y -4) = 0
=> y = -1/2 or 4
log₃x = (-1/2) or 4
=> x = 3^(-1/2) or 3⁴
=> x= 1/√3 or 81
1/√3 - 1 is not greater than 1
=> x = 81
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