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Posted by: Mr. Babatunde« on: February 20, 2020, 12:47:29 PM »The solution to [Mathematics II] Theory Of Logarithms And Indices via our computer base test. 1. Given that log2 = 0.30103 and log3 = 0.4771, calculate Log 6 Log6 = log (2x3) = log2 + log3 = 0.30103 + 0.4771 Ans = 0.77815 2. Given that log2 = 0.30103 and log3 = 0.4771, calculate Log 0.9 Log0.9 = log (9/10) = log9 - log10 = log3^2 - 1 = 2 log 3 - 1 = 2 x 0.4771 - 1 Ans = -0.04576 3. Simplify 2^6 x 2^-6 (Note ^ means Raise to power) 2^6 x 2^-6 = 2^6 + (-6) = 2^6 - 6 = 2^0 = 1 4. Simplify 9^1/6 x 9^1/3 (Note ^ means Raise to power) 9^1/6 x 9^1/3 = 9^1/6 + 1/3 fins the L.c.m = 9^1/2 = √9 Ans = 3 5. Simplify (81/16)^-3/4 (Note ^ means Raise to power) (81/16)^-3/4 = (16/81)^3/4 = (4√16/81)^3 = (2/3)^3 Ans = 8/27 6. Simplify (2d^3)^2 (Note ^ means Raise to power) (2d^3)^2 = 2^2 x (d^3)^2 = 4d^6 7. Simplify 2a^3 ÷ a^4 (Note ^ means Raise to power) 2a^3 ÷ a^4 = 2(a^3 ÷ a^4) = 2a^(3-4) = 2a^-1 Ans = 2/a 8. Simplify 2/5 Log 32 2/5 Log 32 = Log(32)^2/5 = Log (2^5)^2/5 = Log (2)^2 Ans = Log 4 9. Simplify 2 - Log 4 2 - Log 4 = Log (100/4) Ans = Log 25 10. Evaluate Log base3 (27) Log base3 (27) = x then 3^x = 27 3^x = 3^3 Ans x = 3
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