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Given 3^a^2=3^9 & (3^b)^2=3^5 where a>0,b>0 then (2a-b^2)/b-a? (1) 1 (2) 2 (3) 1/2 (4) -2?
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Given 3^a^2=3^9 & (3^b)^2=3^5 where a>0,b>0 then (2a-b^2)/b-a? (1) 1 (...
To solve the equation 3^a^2 = 3^9, we can equate the exponents and solve for a.

We have:
a^2 = 9

Taking the square root of both sides, we get:
a = ± √9

Simplifying, we have:
a = ± 3

Therefore, the solutions for a are a = 3 and a = -3.
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Given 3^a^2=3^9 & (3^b)^2=3^5 where a>0,b>0 then (2a-b^2)/b-a? (1) 1 (2) 2 (3) 1/2 (4) -2?
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Given 3^a^2=3^9 & (3^b)^2=3^5 where a>0,b>0 then (2a-b^2)/b-a? (1) 1 (2) 2 (3) 1/2 (4) -2? for Class 8 2024 is part of Class 8 preparation. The Question and answers have been prepared according to the Class 8 exam syllabus. Information about Given 3^a^2=3^9 & (3^b)^2=3^5 where a>0,b>0 then (2a-b^2)/b-a? (1) 1 (2) 2 (3) 1/2 (4) -2? covers all topics & solutions for Class 8 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Given 3^a^2=3^9 & (3^b)^2=3^5 where a>0,b>0 then (2a-b^2)/b-a? (1) 1 (2) 2 (3) 1/2 (4) -2?.
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