Math Exercises & Math Problems: Logarithmic Equations and Inequalities - Free Printable

To solve the given logarithmic expressions, we will use various properties of logarithms. Here are the solutions for each part:

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a) \(\log 1\)

The logarithm of 1 to any base is always 0.
\[
\log 1 = 0
\]

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b) \(\log_3 27\)

Since \(27 = 3^3\), we have:
\[
\log_3 27 = \log_3 (3^3) = 3
\]

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c) \(\log_4 16\)

Since \(16 = 4^2\), we have:
\[
\log_4 16 = \log_4 (4^2) = 2
\]

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d) \(\log_2 16\)

Since \(16 = 2^4\), we have:
\[
\log_2 16 = \log_2 (2^4) = 4
\]

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e) \(\log_4 2\)

We know that \(2 = 4^{1/2}\), so:
\[
\log_4 2 = \log_4 (4^{1/2}) = \frac{1}{2}
\]

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f) \(\log_{27} 3\)

Since \(27 = 3^3\), we can write \(3\) as \(27^{1/3}\):
\[
\log_{27} 3 = \log_{27} (27^{1/3}) = \frac{1}{3}
\]

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g) \(\log 10,000\)

Since \(10,000 = 10^4\), we have:
\[
\log 10,000 = \log (10^4) = 4
\]

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h) \(\log_3 9^2\)

First, simplify \(9^2\):
\[
9^2 = (3^2)^2 = 3^4
\]
So:
\[
\log_3 9^2 = \log_3 (3^4) = 4
\]

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i) \(\log_{\frac{1}{3}} 9\)

Since \(9 = 3^2\), we can write:
\[
\log_{\frac{1}{3}} 9 = \log_{\frac{1}{3}} (3^2)
\]
Using the change of base property and knowing that \(\frac{1}{3} = 3^{-1}\):
\[
\log_{\frac{1}{3}} (3^2) = \frac{\log_3 (3^2)}{\log_3 (3^{-1})} = \frac{2}{-1} = -2
\]

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j) \(\log_{\frac{1}{3}} \frac{1}{9}\)

Since \(\frac{1}{9} = (3^{-1})^2 = 3^{-2}\), we have:
\[
\log_{\frac{1}{3}} \frac{1}{9} = \log_{\frac{1}{3}} (3^{-2})
\]
Using the change of base property:
\[
\log_{\frac{1}{3}} (3^{-2}) = \frac{\log_3 (3^{-2})}{\log_3 (3^{-1})} = \frac{-2}{-1} = 2
\]

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k) \(\ln e^2\)

The natural logarithm \(\ln\) is the logarithm with base \(e\). So:
\[
\ln e^2 = 2
\]

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l) \(\log_{\frac{1}{5}} \frac{1}{125}\)

Since \(\frac{1}{125} = (5^{-1})^3 = 5^{-3}\), we have:
\[
\log_{\frac{1}{5}} \frac{1}{125} = \log_{\frac{1}{5}} (5^{-3})
\]
Using the change of base property:
\[
\log_{\frac{1}{5}} (5^{-3}) = \frac{\log_5 (5^{-3})}{\log_5 (5^{-1})} = \frac{-3}{-1} = 3
\]

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m) \(\log_{216} 6^{-1}\)

Since \(216 = 6^3\), we have:
\[
\log_{216} 6^{-1} = \log_{6^3} 6^{-1}
\]
Using the change of base property:
\[
\log_{6^3} 6^{-1} = \frac{\log_6 (6^{-1})}{\log_6 (6^3)} = \frac{-1}{3}
\]

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n) \(\log 10^{10}\)

Since \(10^{10}\) is already in the form of a power of 10, we have:
\[
\log 10^{10} = 10
\]

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o) \(\log_8 64\)

Since \(64 = 8^2\), we have:
\[
\log_8 64 = \log_8 (8^2) = 2
\]

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p) \(\log_{\frac{1}{9}} 81\)

Since \(81 = 9^2 = (3^2)^2 = 3^4\) and \(\frac{1}{9} = 3^{-2}\), we have:
\[
\log_{\frac{1}{9}} 81 = \log_{3^{-2}} (3^4)
\]
Using the change of base property:
\[
\log_{3^{-2}} (3^4) = \frac{\log_3 (3^4)}{\log_3 (3^{-2})} = \frac{4}{-2} = -2
\]

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q) \(\log_{21} 21\)

The logarithm of a number to itself is always 1:
\[
\log_{21} 21 = 1
\]

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r) \(\log_{\sqrt{51}} 51^2\)

Since \(51^2 = (\sqrt{51})^4\), we have:
\[
\log_{\sqrt{51}} 51^2 = \log_{\sqrt{51}} ((\sqrt{51})^4) = 4
\]

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s) \(\log_{128} \frac{1}{4}\)

Since \(128 = 2^7\) and \(\frac{1}{4} = 2^{-2}\), we have:
\[
\log_{128} \frac{1}{4} = \log_{2^7} 2^{-2}
\]
Using the change of base property:
\[
\log_{2^7} 2^{-2} = \frac{\log_2 (2^{-2})}{\log_2 (2^7)} = \frac{-2}{7}
\]

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t) \(\log_{\frac{1}{8}} \sqrt{128}\)

Since \(\sqrt{128} = 128^{1/2} = (2^7)^{1/2} = 2^{7/2}\) and \(\frac{1}{8} = 2^{-3}\), we have:
\[
\log_{\frac{1}{8}} \sqrt{128} = \log_{2^{-3}} 2^{7/2}
\]
Using the change of base property:
\[
\log_{2^{-3}} 2^{7/2} = \frac{\log_2 (2^{7/2})}{\log_2 (2^{-3})} = \frac{\frac{7}{2}}{-3} = -\frac{7}{6}
\]

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u) \(\log_3 18 + \log_3 \frac{3}{2}\)

Using the property \(\log_b x + \log_b y = \log_b (xy)\):
\[
\log_3 18 + \log_3 \frac{3}{2} = \log_3 \left(18 \cdot \frac{3}{2}\right) = \log_3 (27) = 3
\]

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v) \(\log e^2 + \log e^{-2}\)

Using the property \(\log_b x + \log_b y = \log_b (xy)\):
\[
\log e^2 + \log e^{-2} = \log (e^2 \cdot e^{-2}) = \log (e^0) = \log 1 = 0
\]

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w) \(\ln e^2 + \ln e^3\)

Using the property \(\ln x + \ln y = \ln (xy)\):
\[
\ln e^2 + \ln e^3 = \ln (e^2 \cdot e^3) = \ln (e^5) = 5
\]

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x) \(\log 20 + \log 50\)

Using the property \(\log x + \log y = \log (xy)\):
\[
\log 20 + \log 50 = \log (20 \cdot 50) = \log 1000 = 3
\]

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y) \(\log 500 - \log 5\)

Using the property \(\log x - \log y = \log \left(\frac{x}{y}\right)\):
\[
\log 500 - \log 5 = \log \left(\frac{500}{5}\right) = \log 100 = 2
\]

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z) \(\log 200 + \log 5 - \log 100\)

Using the properties \(\log x + \log y = \log (xy)\) and \(\log x - \log y = \log \left(\frac{x}{y}\right)\):
\[
\log 200 + \log 5 - \log 100 = \log \left(\frac{200 \cdot 5}{100}\right) = \log 10 = 1
\]

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Z) \(\log_5 20 + \log_5 \frac{125}{4} - \log_5 5^{-1}\)

Using the properties \(\log_b x + \log_b y = \log_b (xy)\) and \(\log_b x - \log_b y = \log_b \left(\frac{x}{y}\right)\):
\[
\log_5 20 + \log_5 \frac{125}{4} - \log_5 5^{-1} = \log_5 \left(20 \cdot \frac{125}{4} \cdot 5\right)
\]
Simplify the expression inside the logarithm:
\[
20 \cdot \frac{125}{4} \cdot 5 = \frac{20 \cdot 125 \cdot 5}{4} = \frac{12500}{4} = 3125
\]
So:
\[
\log_5 3125 = \log_5 (5^5) = 5
\]

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Final Answers:

\[
\boxed{
\begin{array}{lll}
a) & 0 & j) & 2 & s) & -\frac{2}{7} \\
b) & 3 & k) & 2 & t) & -\frac{7}{6} \\
c) & 2 & l) & 3 & u) & 3 \\
d) & 4 & m) & -\frac{1}{3} & v) & 0 \\
e) & \frac{1}{2} & n) & 10 & w) & 5 \\
f) & \frac{1}{3} & o) & 2 & x) & 3 \\
g) & 4 & p) & -2 & y) & 2 \\
h) & 4 & q) & 1 & z) & 1 \\
i) & -2 & r) & 4 & Z) & 5 \\
\end{array}
}
\]