Solution 3.3:4c

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All three arguments of the logarithm can be written as powers of
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All three arguments of the logarithm can be written as powers of 3,
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<math>\text{3}</math>,
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& 27^{\frac{1}{3}}=\left( 3^{3} \right)^{\frac{1}{3}}=3^{3\centerdot \frac{1}{3}}=3^{1}=3, \\
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27^{\frac{1}{3}} &= \bigl(3^3\bigr)^{\frac{1}{3}} = 3^{3\cdot\frac{1}{3}} = 3^1 = 3\,,\\[5pt]
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& \frac{1}{9}=\frac{1}{3^{2}}=3^{-2} \\
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\frac{1}{9} &= \frac{1}{3^2} = 3^{-2}\,,\\
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\end{align}</math>
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\end{align}</math>}}
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and it is therefore appropriate to use base 3 when simplifying using the logarithms, even if we have the base 10-logarithm, lg,
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{{Displayed math||<math>\begin{align}
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and it is therefore appropriate to use base
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\lg 27^{\frac{1}{3}} + \frac{\lg 3}{2} + \lg \frac{1}{9}
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<math>\text{3}</math>
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&= \lg 3 + \frac{1}{2}\lg 3 + \lg 3^{-2}\\[5pt]
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when simplifying using the logarithms, even if we have the base
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&= \lg 3 + \frac{1}{2}\lg 3 + (-2)\cdot\lg 3\\[5pt]
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<math>\text{1}0</math>
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&= \Bigl(1+\frac{1}{2}-2\Bigr)\lg 3\\[5pt]
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logarithm, lg,
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&= -\frac{1}{2}\lg 3\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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& \lg 27^{\frac{1}{3}}+\frac{\lg 3}{2}+\lg \frac{1}{9}=\lg 3+\frac{1}{2}\lg 3+\lg 3^{-2} \\
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& =\lg 3+\frac{1}{2}\lg 3+\left( -2 \right)\centerdot \lg 3 \\
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& =\left( 1+\frac{1}{2}-2 \right)\lg 3=-\frac{1}{2}\lg 3 \\
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\end{align}</math>
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This expression cannot be simplified any further.
This expression cannot be simplified any further.

Current revision

All three arguments of the logarithm can be written as powers of 3,

\displaystyle \begin{align}

27^{\frac{1}{3}} &= \bigl(3^3\bigr)^{\frac{1}{3}} = 3^{3\cdot\frac{1}{3}} = 3^1 = 3\,,\\[5pt] \frac{1}{9} &= \frac{1}{3^2} = 3^{-2}\,,\\ \end{align}

and it is therefore appropriate to use base 3 when simplifying using the logarithms, even if we have the base 10-logarithm, lg,

\displaystyle \begin{align}

\lg 27^{\frac{1}{3}} + \frac{\lg 3}{2} + \lg \frac{1}{9} &= \lg 3 + \frac{1}{2}\lg 3 + \lg 3^{-2}\\[5pt] &= \lg 3 + \frac{1}{2}\lg 3 + (-2)\cdot\lg 3\\[5pt] &= \Bigl(1+\frac{1}{2}-2\Bigr)\lg 3\\[5pt] &= -\frac{1}{2}\lg 3\,\textrm{.} \end{align}

This expression cannot be simplified any further.

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