The number of digits of $a^b=\lfloor{1+b*{log_{10}}(a)}\rfloor$ So $m=1+1997 \cdot {log_{10}}(2)=602$ And $n=1+1997 \cdot {log_{10}}(5)=1396$ So $m+n=1396+602=1998$

@Above you can use \cdot or \times

Bratin_Dasgupta wrote: @Above you can use \cdot or \times Ok,thanks.

@3above, Just to clarify, the formula you presented is slightly off, instead, you have to take the floor function of the $\log_{10}{n}$ as that value isn't always an integer.

NealShrestha wrote: @3above, Just to clarify, the formula you presented is slightly off, instead, you have to take the floor function of the $\log_{10}{n}$ as that value isn't always an integer. Yeah

The answer is 1998 Just see the product and show that the sum of the digits of 2 factors of 10^k is the number of digits of 10^k

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Sprites wrote: The number of digits of $a^b=\lfloor{1+b*{log_{10}}(a)}\rfloor$ So $m=1+1997 \cdot {log_{10}}(2)=602$ And $n=1+1997 \cdot {log_{10}}(5)=1396$ So $m+n=1396+602=1998$ for $b>0$ but for $b<0$ ? example for $2^{-n}$ n-positiv integernumber.

Note that $10^{m-1} < 2^{1997} < 10^m$ and $10^{n-1} < 5^{1997} < 10^n$. Multiplying these two together yields $10^{m+n-2} < 10^{1997} < 10^{m+n}$ from which it follows that $m+n=1998$.

Aiden-1089 wrote: Note that $10^{m-1} < 2^{1997} < 10^m$ and $10^{n-1} < 5^{1997} < 10^n$. Multiplying these two together yields $10^{m+n-2} < 10^{1997} < 10^{m+n}$ from which it follows that $m+n=1998$. Thanks very much, but for $2^{-n}$ n-positiv integer number. How many decimalplases exist?

without logaritm is hard to prove

why answer is $n$ ?