Evaluating Logarithms 3
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Question 1 of 4
1. Question
Solve for `log_e 7.5`, given that:`log_e 2=0.6931``log_e 3=1.0986``log_e 5=1.6094`Round your answer to 4 decimal places- `log_e 7.5=` (2.0149)
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Laws of Logarithms
$$\log_{\color{#9a00c7}{b}} {\color{#00880A}{x}\color{#e65021}{y}}=\log_{\color{#9a00c7}{b}} \color{#00880A}{x} + \log_{\color{#9a00c7}{b}} \color{#e65021}{y}$$$$\log_{\color{#9a00c7}{b}} \frac{\color{#00880A}{x}}{\color{#e65021}{y}}=\log_{\color{#9a00c7}{b}} \color{#00880A}{x}-\log_{\color{#9a00c7}{b}} \color{#e65021}{y}$$Expand the given logarithmic expression$$\log_e 7.5$$ `=` $$\log_e \frac{15}{2}$$ `7.5=15/2` `=` $$\log_e \frac{(5)(3)}{2}$$ `15=(5)(3)` `=` $$\log_\color{#9a00c7}{e} {\frac{\color{#00880A}{(5)(3)}}{\color{#e65021}{2}}}$$ `=` $$\log_\color{#9a00c7}{e} \color{#00880A}{(5)(3)}-\log_\color{#9a00c7}{e} \color{#e65021}{2}$$ $$log_b \frac{x}{y}=log_b x-\log_b y$$ `=` $$\log_\color{#9a00c7}{e} \color{#00880A}{(5)}\color{#e65021}{(3)}-\log_e 2$$ `=` $$\log_\color{#9a00c7}{e} \color{#00880A}{5}+\log_\color{#9a00c7}{e} \color{#e65021}{3}-\log_e 2$$ `log_b xy=log_b x+log_b y` Substitute the given values$$\log_e 2$$ `=` $$0.6931$$ $$\log_e 3$$ `=` $$1.0986$$ $$\log_e 5$$ `=` $$1.6094$$ `log_e 5+log_e 3-log_e 2` `=` `1.6094+1.0986-0.6931` `=` `2.0149` `2.0149` -
Question 2 of 4
2. Question
Solve for `log_e 200`, given that:`log_e 2=0.6931``log_e 5=1.6094`Round your answer to 4 decimal places- `log_e 200=` (5.2981)
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Laws of Logarithms
$$\log_{\color{#9a00c7}{b}} {\color{#00880A}{x}\color{#e65021}{y}}=\log_{\color{#9a00c7}{b}} \color{#00880A}{x} + \log_{\color{#9a00c7}{b}} \color{#e65021}{y}$$$$\log_b x^\color{#004ec4}{p}=\color{#004ec4}{p}\log_b x$$Expand the given logarithmic expression$$\log_e 200$$ `=` $$\log_e (25)(8)$$ `200=(25)(8)` `=` $$\log_\color{#9a00c7}{e} \color{#00880A}{(25)}\color{#e65021}{(8)}$$ `=` $$\log_\color{#9a00c7}{e} \color{#00880A}{25}+\log_\color{#9a00c7}{e} \color{#e65021}{8}$$ `log_b xy=log_b x+log_b y` `=` $$\log_e 5^2+\log_e 8$$ `25=5^2` `=` $$\log_e 5^2+\log_e 2^3$$ `8=2^3` `=` $$\color{#004ec4}{2}\log_{e} 5+\color{#004ec4}{3}\log_{e} 2$$ `log_b x^p=p log_b x` Substitute the given values$$\log_e 2$$ `=` $$0.6931$$ $$\log_e 5$$ `=` $$1.6094$$ `2log_e 5+3log_e 2` `=` `2(1.6094)+3(0.6931)` `=` `3.2188+2.0793` `=` `5.2981` `5.2981` -
Question 3 of 4
3. Question
Solve for `log_e (1/25)`, given that:`log_e 5=1.6094`Round your answer to 4 decimal places- `log_e (1/25)=` (-3.2188)
Hint
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Laws of Logarithms
$$\log_b x^\color{#004ec4}{p}=\color{#004ec4}{p}\log_b x$$Simplify the given logarithmic expression$$\log_e \frac{1}{25}$$ `=` $$\log_e \frac{1}{\color{#CC0000}{5^2}}$$ `25=5^2` `=` $$\log_e 5^{-2}$$ Reciprocate `1/(5^2)` `=` $$\color{#004ec4}{-2}\log_{e} 5$$ `log_b x^p=p log_b x` Substitute the given values`log_e 5` `=` `1.6094` `-2log_e 5` `=` `-2(1.6094)` `=` `-3.2188` `-3.2188` -
Question 4 of 4
4. Question
Solve for `log_e 2.5`, given that:`log_e 2=0.6931``log_e 5=1.6094`Round your answer to 4 decimal places- `log_e 2.5=` (0.9163)
Hint
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Exceptional!
Incorrect
Laws of Logarithms
$$\log_{\color{#9a00c7}{b}} \frac{\color{#00880A}{x}}{\color{#e65021}{y}}=\log_{\color{#9a00c7}{b}} \color{#00880A}{x}-\log_{\color{#9a00c7}{b}} \color{#e65021}{y}$$Expand the given logarithmic expression$$\log_e 2.5$$ `=` $$\log_e \frac{5}{2}$$ `2.5=5/2` `=` $$\log_\color{#9a00c7}{e} {\frac{\color{#00880A}{5}}{\color{#e65021}{2}}}$$ `=` $$\log_\color{#9a00c7}{e} \color{#00880A}{5}-\log_\color{#9a00c7}{e} \color{#e65021}{2}$$ $$log_b \frac{x}{y}=log_b x-\log_b y$$ Substitute the given values$$\log_e 2$$ `=` $$0.6931$$ $$\log_e 5$$ `=` $$1.6094$$ `log_e 5-log_e 2` `=` `1.6094-0.6931` `=` `0.9163` `0.9163`
Quizzes
- Converting Between Logarithmic and Exponent Form 1
- Converting Between Logarithmic and Exponent Form 2
- Evaluating Logarithms 1
- Evaluating Logarithms 2
- Evaluating Logarithms 3
- Expanding Log Expressions
- Simplifying Log Expressions 1
- Simplifying Log Expressions 2
- Simplifying Log Expressions 3
- Change Of Base Formula
- Logarithmic Equations 1
- Logarithmic Equations 2
- Logarithmic Equations 3
- Solving Exponential Equations