If log_{10} 2 = 0.3010 and log_{10} 3 = 0.477 | vedclass

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(B) Let $x = (0.0432)^{10}$.Taking $\log_{10}$ on both sides:$\log_{10} x = 10 \log_{10} (0.0432) = 10 \log_{10} (432 	imes 10^{-4})$$= 10 [\log_{10}

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) Let $x = (0.0432)^{10}$.Taking $\log_{10}$ on both sides:$\log_{10} x = 10 \log_{10} (0.0432) = 10 \log_{10} (432 	imes 10^{-4})$$= 10 [\log_{10} (2^4 	imes 3^3) - 4]$$= 10 [4 \log_{10} 2 + 3 \log_{10} 3 - 4]$$= 10 [4(0.3010) + 3(0.4771) - 4]$$= 10 [1.2040 + 1.4313 - 4]$$= 10 [2.6353 - 4] = 10 [-1.3647] = -13.647$To express this in standard form,we write: $-13.647 = -14 + 0.353 = \bar{14}.353$.The characteristic is $-14$. The number of zeros after the decimal point before the first significant figure is given by $|	ext{characteristic}| - 1 = 14 - 1 = 13$.

If $\log_{10} 2 = 0.3010$ and $\log_{10} 3 = 0.4771$,then the number of zeros between the decimal point and the first significant figure in $(0.0432)^{10}$ is:

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