If 1, log9 (3^(1–x) + 2) and log3 (4.3^x –1) are in A.P., then x is equal to -Maths 9th

If 1, log9 (3^(1–x) + 2) and log3 (4.3^x –1) are in A.P., then x is equal to -Maths 9th
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(d) log3\(\big(rac{3}{4}\big)\) 1, log9 (31 – x + 2), log3 (4.3x –1) are in A.P. ⇒ log33 , log3 (31– x + 2)1/2, log3 (4.3x – 1) are in A.P. ⇒ 3, (31 – x + 2)1/2, (4.3x – 1) are in G.P.\(\bigg[\)Since log9 (31 – x + 2) = log32 (31 – x + 2) = \(rac{1}{2}\) log3 (31 – x + 2), using logan \(x\) = \(rac{1}{n}\)loga \(x\) = log3 (31 – x + 2)\(^{rac{1}{2}}\)\(\bigg]\)⇒ [[31–x + 2]1/2]2 = 3.(4.3x – 1) ⇒ 31 – x + 2 = 4.3x + 1 – 3⇒ 4.3x + 1 – 31 - x = 5 ⇒ 12.3x - \(rac{3}{3^x}\) = 5Let 3x = y, then 12y – \(rac{3}{y}\) = 5 ⇒ 12y2 – 5y –3 = 0⇒ (3y + 1) (4y – 3) = 0 ⇒ y = - \(rac{1}{3}\), \(rac{3}{4}\)∴ Rejecting the negative value, we have 3x = \(rac{3}{4}\)⇒ x = log3\(rac{3}{4}.\)
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