MATH116 Lecture Notes - Lecture 14: Opata Language, National Council Of Educational Research And Training

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Practice solutions 3: (a) solve log5 (x + 4) + log5 (x) = 1 for x. (b) solve 4x + 2x+1 = 18 for x. (cid:18)3x 2 (cid:19) x 1 (c) solve ln. < 0. log5 (x + 4) + log5 (x) = 1 log5 (x(x + 4)) = 1 x2 + 4x = 5 x2 + 4x 5 = 0 (x + 5)(x 1) = 0. We see that x = 5 and x = 1 are possible solutions but neither log5 (x + 4) nor log5 (x) are de ned for x = 5. Thus, the only solution is x = 1. (b) since 4 = 22 and 2x+1 = 2x21, we can write this equation as (22)x + (2)2x = 18. We can think of this as a quadratic in the variable q = 2x, that is q2 + 2q 18 = 0. Using the quadratic equation, we have that q =

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