X = 2.35 x 10-5. Therefore -log(X) = _____ to the correct number of significant figures: 4.6289 4.620 4.6 4.629 none of the above is correct
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-log(X) = -log(2.35 × 10^-5) Show more…
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Solve the equation $x^{3.2}=41.15$ correct to 4 significant figures. Taking logarithms to base 10 of both sides gives: $\log _{10} x^{3.2}=\log _{10} 41.15$ $3.2 \log _{10} x=\log _{10} 41.15$ Hence $\log _{10} x=\frac{\log _{10} 41.15}{3.2}=0.50449$ Thus $x=$ antilog $0.50449=10^{0.50449}=\mathbf{3 . 1 9 5}$ correct to 4 significant figures.
Correct any answers that have the incorrect number of significant figures. (a) $\left(3.8 \times 10^{5}\right)-\left(8.45 \times 10^{5}\right)=-4.7 \times 10^{5}$ (b) $0.00456+1.0936=1.10$ (c) $8475.45-34.899=8440.55$ (d) $908.87-905.34095=3.5291$
Correct any answers that have the incorrect number of significant figures. (a) $34.00 \times 567 \div 4.564=4.2239 \times 10^{3}$ (b) $79.3 \div 0.004 \times 35.4=7 \times 10^{5}$ (c) $89.763 \div 22.4581=3.997$ (d) $\left(4.32 \times 10^{12}\right) \div\left(3.1 \times 10^{-4}\right)=1.4 \times 10^{16}$
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