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
