maths question... solution needed
Answers
Question :-
If log₃2, log₃(2^x - 5), log₃(2^x - 7/2) are in AP. Then find the value of x.
Answer :-
x = 3
Solution :-
log₃2, log₃(2^x - 5), log₃(2^x - 7/2) are in AP
Here,
- a = log₃2
- a2 = log₃(2^x - 5)
- a3 = log₃(2^x - 7/2)
Since, they are in AP,
==> Common difference = a2 - a = a3 - a2
==> log₃(2^x - 5) - log₃2 = log(2^x - 7/2) - log₃(2^x - 5)
==> log₃(2^x - 5) + log₃(2^x - 5) = log₃2 + log₃(2^x - 7/2)
Using Product rule log a + log b = log ab
==> 2log₃(2^x - 5) = log₃2(2^x - 7/2)
Using Power rule m.log a = log a^m
==> log₃(2^x - 5)² = log₃2(2^x - 7/2)
Comparing on both sides
==> (2^x - 5)² = 2(2^x - 7/2)
Substituting 2^x = y in the above equation
==> (y - 5)² = 2(y - 7/2)
==> y² - 2( y )( 5 ) + 5² = 2y - 7
==> y² - 10y + 25 = 2y - 7
==> y² - 10y - 2y + 25 + 7 = 0
==> y² - 12y + 32 = 0
==> y² - 8y - 4y + 32 = 0
==> y(y - 8) - 4(y - 8) = 0
==> (y - 4)(y - 8) = 0
==> y - 4 = 0 or y - 8 = 0
==> y = 4 or y = 8
==> 2^x = 2² or 2^x = 2³
Since bases are equal we can equate powers
==> x = 2 or x = 3.
x ≠ 2, because it is not possible according to the data
==> x = 3
Therefore the value of x is 3.
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