math ce

1)Since f(-1) * f(0) < 0 and f(1) * f(2) < 0 , the roots of f(x) = 0
between - 1 and 0 or between 1 and 2.Let - 1 < α < 0 and 1 < β < 2 be the roots of x^2 + bx + c = 0 Sum of the roots : α + β = - b
Product of the roots : α β = c- 1 + 1 < α + β < 0 + 2
0 < - b < 2
- 2 < b < 0 (The range of b)- 1 * 2 < α β < 0
- 2 < c < 0 (The range of c)
2)f(x^5) = logxLet x^5 = 2 , then
5logx = log2
logx = (1/5)log2
i.e.
f(x^5) = (1/5)log2
f(2) = (1/5)log2
3)2^8 * 8^10 * 5^12= 2^8 * (2^3)^10 * 5^12
= 2^38 * 5^12
= (2*5)^12 * 2^26
= 2^26 * 10^12
= 67,108,864,000,000,000,000 is a 20 digits number.
4)log (1 - 1/9) = a
log (8/9) = a
log 8 - log 9 = a ....(1)log (1 - 1/81) = b
log (80/81) = b
log 80 - log 81 = b
log 80 - 2log 9 = b ....(2)(2) - (1)*2 :log 80 - 2log 9 - (2log 8 - 2log 9) = b - 2a
log 80 - 2log 8 = b - 2a
(log 10 + log 8) - 2log 8 = b - 2a
1 - log 8 = b - 2a
log 8 = 2a - b + 1
3log 2 = 2a - b + 1
log 2 = (1/3) (2a - b + 1)

2. 1/5 log 2