If ${log}_{10} 2 = 0.30103,$ determine how many zeros are there between the decimal point and the first significant digit in ${(\frac{1}{2})}^{1000}$ .

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Solution

Let, $x = {(\frac{1}{2})}^{1000} .$
Then, ${log}_{10} x = - 1000 log 2 = - 1000 \times .30103$
$= - 301.03 = ¯ ¯¯¯¯¯¯ ¯ 302 .97 .$
Hence, the number of zeros in ${(\frac{1}{2})}^{1000}$ between the decimal point and the first significant digit is $(302 - 1) = 301.$
Ans: 301

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If log102=0.30103, determine how many zeros are there between the decimal point and the first significant digit in (12)1000.

Let, x=(12)1000.Then, log10x=−1000log2=−1000×.30103=−301.03=¯¯¯¯¯¯¯¯302.97.Hence, the number of zeros in (12)1000 between the decimal point and the first significant digit is (302−1)=301. Ans: 301

If ${log}_{10} 2 = 0.30103,$ determine how many zeros are there between the decimal point and the first significant digit in ${(\frac{1}{2})}^{1000}$ .

Let, $x = {(\frac{1}{2})}^{1000} .$
Then, ${log}_{10} x = - 1000 log 2 = - 1000 \times .30103$
$= - 301.03 = ¯ ¯¯¯¯¯¯ ¯ 302 .97 .$
Hence, the number of zeros in ${(\frac{1}{2})}^{1000}$ between the decimal point and the first significant digit is $(302 - 1) = 301.$
Ans: 301