Ans. 8
[ ^ = raised to]
Given :
a = (1/9)^-2log₃7
b = 2^ -log₁/₂7
Now "a" can be written as :
a = (1/3²)^ -2log₃7
= (3⁻²)^ -2log₃7
= 3^⁴log₃7 [∵ -2 × -2 = 4]
Now 4 will go to power of 7.
= 3^log₃7⁴
But 3^log₃ = 1 [∵ a^logₐ = 1]
∴ [a = 7⁴]
Now "b" can be written as:
b = 2^ -log(₂⁻1)7 [ ∵ 1/a = a⁻1]
Now -1 will go from down ahead of log to become denominator of the ahead term i.e. [ logₐ^x = 1/xlogₐ ]
= 2^log₂7
But 2^log₂ = 1 [∵ a^logₐ = 1]
∴ b = 7
Now by comparing "a" and "b"
it is given :
a = b^(k)
Now "a" = 7⁴
And "b" = 7
∴ 7⁴ = 7^(k)
Hence by comparing the exponents [k = 4].
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