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Math  /  Algebra

QuestionA) 0.0659 B) 0.1505 C) 7.5820 D) 15.1639
1 the value. Give an approximation to four decimal places. 2) log⁡563−log⁡291\log 563-\log 291 A) 12.0066 B) 1.1163 C) 0.6600 3) ln⁡115−ln⁡24\ln 115-\ln 24 A) 7.9230 B) 1.4930 D) 0.2866 D) -5.0158
Ive the problem. 4) Let u=ln⁡au=\ln a and v=ln⁡bv=\ln b. Write the following expression in terms of uu and vv without using⁡u \operatorname{sing} the function ln⁡\ln. ln⁡(ab85)\ln \left(\sqrt[5]{a b^{8}}\right) A) u5−8v\frac{u}{5}-8 v B) u5+85v\frac{u}{5}+\frac{8}{5} v C) 85u+85v\frac{8}{5} u+\frac{8}{5} v D) u5−85v\frac{u}{5}-\frac{8}{5} v 5) Let u=ln⁡a\mathrm{u}=\ln \mathrm{a} and v=ln⁡b\mathrm{v}=\ln \mathrm{b}. Write the following expression in terms of u and v without using the function In. ln⁡a9b35\ln \sqrt[5]{\frac{a^{9}}{b^{3}}}

Studdy Solution

STEP 1

1. We are dealing with logarithmic expressions and need to simplify them.
2. We will use logarithmic properties to simplify expressions.
3. We will approximate numerical values to four decimal places where necessary.

STEP 2

1. Simplify log⁡563−log⁡291\log 563 - \log 291 using logarithmic properties.
2. Simplify ln⁡115−ln⁡24\ln 115 - \ln 24 using logarithmic properties.
3. Express ln⁡(ab85)\ln \left(\sqrt[5]{a b^{8}}\right) in terms of uu and vv.
4. Express ln⁡a9b35\ln \sqrt[5]{\frac{a^{9}}{b^{3}}} in terms of uu and vv.

STEP 3

Use the property of logarithms: log⁡a−log⁡b=log⁡(ab)\log a - \log b = \log \left(\frac{a}{b}\right).
log⁡563−log⁡291=log⁡(563291) \log 563 - \log 291 = \log \left(\frac{563}{291}\right)
Calculate 563291\frac{563}{291} and then find the logarithm.
563291≈1.9347 \frac{563}{291} \approx 1.9347
Now calculate log⁡(1.9347)\log(1.9347).
log⁡(1.9347)≈0.2866 \log(1.9347) \approx 0.2866

STEP 4

Use the property of logarithms: ln⁡a−ln⁡b=ln⁡(ab)\ln a - \ln b = \ln \left(\frac{a}{b}\right).
ln⁡115−ln⁡24=ln⁡(11524) \ln 115 - \ln 24 = \ln \left(\frac{115}{24}\right)
Calculate 11524\frac{115}{24} and then find the natural logarithm.
11524≈4.7917 \frac{115}{24} \approx 4.7917
Now calculate ln⁡(4.7917)\ln(4.7917).
ln⁡(4.7917)≈1.5630 \ln(4.7917) \approx 1.5630

STEP 5

Use the properties of logarithms to express ln⁡(ab85)\ln \left(\sqrt[5]{a b^{8}}\right) in terms of uu and vv.
ln⁡(ab85)=ln⁡((ab8)1/5) \ln \left(\sqrt[5]{a b^{8}}\right) = \ln \left((a b^{8})^{1/5}\right)
=15ln⁡(ab8) = \frac{1}{5} \ln (a b^{8})
=15(ln⁡a+ln⁡b8) = \frac{1}{5} (\ln a + \ln b^{8})
=15(u+8v) = \frac{1}{5} (u + 8v)
=u5+8v5 = \frac{u}{5} + \frac{8v}{5}

STEP 6

Use the properties of logarithms to express ln⁡a9b35\ln \sqrt[5]{\frac{a^{9}}{b^{3}}} in terms of uu and vv.
ln⁡a9b35=ln⁡((a9b3)1/5) \ln \sqrt[5]{\frac{a^{9}}{b^{3}}} = \ln \left(\left(\frac{a^{9}}{b^{3}}\right)^{1/5}\right)
=15ln⁡(a9b3) = \frac{1}{5} \ln \left(\frac{a^{9}}{b^{3}}\right)
=15(ln⁡a9−ln⁡b3) = \frac{1}{5} (\ln a^{9} - \ln b^{3})
=15(9ln⁡a−3ln⁡b) = \frac{1}{5} (9\ln a - 3\ln b)
=9u5−3v5 = \frac{9u}{5} - \frac{3v}{5}
The solutions are: 1) log⁡563−log⁡291≈0.2866\log 563 - \log 291 \approx 0.2866 2) ln⁡115−ln⁡24≈1.5630\ln 115 - \ln 24 \approx 1.5630 3) ln⁡(ab85)=u5+8v5\ln \left(\sqrt[5]{a b^{8}}\right) = \frac{u}{5} + \frac{8v}{5} 4) ln⁡a9b35=9u5−3v5\ln \sqrt[5]{\frac{a^{9}}{b^{3}}} = \frac{9u}{5} - \frac{3v}{5}

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