Form4 maths question about logarithmic functions(2)

36.
9^(1/h) = 99
(1/h)log 9 = log 99
h = log 9/log 99.
11^(1/k) = 99
(1/k) log 11 = log 99
k = log 11/log 99
So h + k = log 9/log 99 + log 11/log 99 = (log 9 + log 11)/log 99 = log 99/log 99 = 1.
37.
(a) f(1) = 100^log 1 = 100^0 = 1.
(b) f(2) = 100^log 2 = 10^(2log 2) = 10^(log 4) = 4.
f(3) = 100^log 3 = 10^(2log 3) = 10^(log 9) = 9.
So f(2) + f(3) = 4 + 9 = 13.
39.
(k-1)^3 > 2005
3log( k-1) > log 2005
3log(k-1)> 3.302
log (k -1) > 1.1007
k -1 > 12.61
k > 13.61
so k = 14 is the smallest integer.
38.
(sqrt 179 - 13)^20 times (sqrt 179 + 13)^20
= [(sqrt 179 - 13)(sqrt 179 + 13)]^20 = [179 - 13^2]^20
= [179 - 169]^20 = 10^20.
So log 10^20 = 20 log 10 = 20.

36. 9^(1/h)=99=9*11
=>(1/h)log9=log9+log11
=>log9=hlog9+hlog11...(1)

11^(1/k)=99=9*11
=>(1/k)log11=log9+log11
=>log11=klog9+klog11...(2)
(1)+(2) gives log9+log11=(h+k)(log9+log11)
=> h+k=1

37a. f(1)=100^(log1)=100^0=1
37b. f(2)+f(3)=100^(log2)+100^(log3)
= 10^(2log2)+10^(2log3)
= 10^(log4)+10^(log9)
= 4+9 (since 10^(logk)=k)
= 13

38. log[(√179 -13)^20 (√179 +13)^20]
= 20log[(√179 -13)(√179 +13)]
= 20log(179 -13^2)
=20log10
=20

39. (k-1)^3>2005
k>2005^(1/3)+1=13.61
So smallest integer value of k is 14.