9log31 - 2x = 5x2- 5⇒ log3 9log31 - 2x = log35x2- 5⇒ log31 - 2x × log39= log35x2- 5⇒ log31 - 2x × log332 = log35x2- 5⇒ 2log31 - 2x = log35x2- 5 as, log332 = 2log33 = 2 × 1 = 2⇒ log31 - 2x2 = log35x2- 5⇒ 1 - 2x2 = 5x2- 5 ⇒1 + 4x2 - 4x = 5x2- 5⇒ x2 + 4x - 6 = 0a = 1 ; b = 4 ; c = -6Now, D = b2 - 4ac= 42 - 4 × 1 × -6= 16 + 24= 40So, x = -4 ± 402 = -4 ± 2102⇒x = -10 + 2 and x = 10 - 2 rejectedAs log31 - 2x = log31 - 210 + 4 = log35-210 = log3a negative quantity = not definedSo, x = -10 + 2

plz ans this

plz ans this log. (i — 2x) 5 $2 — 5.

9^{\log_{3} (1 - 2 x)} = 5 x^{2} - 5 \Rightarrow \log_{3} [9^{\log_{3} (1 - 2 x)}] = \log_{3} (5 x^{2} - 5) \Rightarrow \log_{3} (1 - 2 x) \times \log_{3} 9 = \log_{3} (5 x^{2} - 5) \Rightarrow \log_{3} (1 - 2 x) \times \log_{3} 3^{2} = \log_{3} (5 x^{2} - 5) \Rightarrow 2 \log_{3} (1 - 2 x) = \log_{3} (5 x^{2} - 5) [as, \log_{3} 3^{2} = 2 \log_{3} 3 = 2 \times 1 = 2] \Rightarrow \log_{3} {(1 - 2 x)}^{2} = \log_{3} (5 x^{2} - 5) \Rightarrow {(1 - 2 x)}^{2} = (5 x^{2} - 5) \Rightarrow 1 + 4 x^{2} - 4 x = 5 x^{2} - 5 \Rightarrow x^{2} + 4 x - 6 = 0 a = 1; b = 4; c = - 6 Now, D = b^{2} - 4 ac = {(4)}^{2} - 4 \times 1 \times (- 6) = 16 + 24 = 40 So, x = \frac{- 4 \pm \sqrt{40}}{2} = \frac{- 4 \pm 2 \sqrt{10}}{2} \Rightarrow x = - (\sqrt{10} + 2) and x = \sqrt{10} - 2 (rejected) As \log_{3} (1 - 2 x) = \log_{3} [1 - 2 \sqrt{10} + 4] = \log_{3} (5 - 2 \sqrt{10}) = \log_{3} (a negative quantity) = not defined So, x = - (\sqrt{10} + 2)

View Full Answer

plz ans this plz ans this log. (i — 2x) 5 $2 — 5.

plz ans this

plz ans this log. (i — 2x) 5 $2 — 5.