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Step 8: Its inverse is (the square root or ( )1/2, the power half ). Step 9: Applying this inverse to both sides of the equation we get x+ 2 b b 2 − 4ac x + b = ± b − 4ac or 1,2 = 2a 2a 2a 2a The latter notation means that if we choose the top sign in the RHS, we get solution x1 and if we choose the bottom sign, we get solution x2. Step 10: The unknown is not the subject of the last equation, we have to go through the solution loop again. The last operation on x1,2 in the LHS is + b . 2a Step 11: Its inverse is − b .

Evaluate and check check your your result: result: b) 9 3. Evaluate and 4 a) c) 416 3. a) Evaluate and check your result: b) 925 d) b) 9 4 a) c) 16 1/ 2 c) e) 36 b) 916 d) 25 25 1/ 2 d) f) 49 c) 16 1 /2 11/ 3/ 2 e) e) g) 36 836 d) 25 11// 22 f) 49 f) 49 h) 36 2711/ /23 e) 1//33 g) 1/ 2 g) 883164 i) 49 f) 1/ 3 327 h) 1125 /13/ 3 h) j) 827 g) 33 64 64 1/ 3 i)i) 27 h) 3 4. j)Evaluate without a calculator: 3 125 j) 3 64 125 i) a) log10 2 log10 5 3 125 j) 4. b) Evaluate without calculator: log10 200 log10aa2calculator: 4.

Logb xn = n logb x log c a 7. 5 Variables: Irrational, Real and Complex Numbers The operation of taking a log is the fourth inverse operation we encounter. The operation of taking a log is the fourth inverse operation we encounter. Question: what other inverse operations have we covered so far? Answer: what other inverse operations have we covered so far? Question: Answer: As with subtraction, division and taking the whole root, application of a log might cause a difficulty: Taking a log of a rational number does not always result in a rational number and taking a log of a real number does not always result in a real number.

### Elementary Algebra and Calculus by Bookboon.com

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