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Find the number of digits in integral part of $60^{12} + 60^{- 12} - 60^{- 15}$ .

(Given log 2 = 0.3030, log 3 = 0.4771)

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Detailed Solution

60^{- 12} = \frac{1}{60^{12}} and 60^{- 15} = \frac{1}{60^{15}}

Both of these are way less than zero and hence these terms are only responsible for the decimal part

Only 60^{12} is responsible for the integral part

Le t^{'} s assume x = 60^{12}

\log log x = \log \log 60^{12}

\log log x = 12 \log \log 60

\log log x = 12 \times \log \log (3 \times 2 \times 10)

\log log x = 12 [\log \log 3 + \log \log 2 + \log \log 10]

\log log x = 12 [0.4771 + 0.3030 + 1]

\log log x = 21.3612

x = 10^{21.3612}

Power of 10 implies that

many zeros followed after 1 and decimal numbers won' t affect that number . This implies 60^{12} has 21 + 1 = 22 digits in its integral part

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