Functions and Logarithms

$h_1(x)=g(f(x)) = \log\left(\dfrac{f(x)}{10}\right)= \log\left(\dfrac{10^{10x}}{10}\right) = 10x-1$

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Thus, we have $h_1(x)=10x-1$ , and as $h_n=h_1(h_{n-1}(x))$ , we have

$h_n=10\times h_{n-1}(x) -1$

Thus we have a recurrence, so now, we'll substitute the value of $x$ as $1$ .

$h_1(1)=10-1=9$ . Hence we have a recursive formula as

$a_n=10a_{n-1}-1$ with given that $a_1=9$ . (Here $a_{n}=h_n(1)$ )

Computing the terms, simply, we have

$a_1=10\times 1-1= 9 \\a_2=10\times 9 -1 =89 \\ a_3=10\times 89 -1 = 889 \\ a_4=10\times 889-1 = 8889 \\ ........$

Now observe that $a_n$ is simply a $n-digit$ number, which has last digit $9$ and all other digits are $8$ .

Hereby, we conclude that $a_{2011} = h_{2011}(1) = \underbrace{888.....888}_\text{2010 times digit '8'}9$

Hence digit sum is $2010\times 8+9 = 16080+9=\boxed{16089}$

h_(n) (x) =10^n x - (n-1)1's. Thus, for n = 2011 => 10^2011 - (2011)1's = (2010)8's and last a 9. Thus, 2010 8 +9 =16089