Logarithmic Equations 3
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Question 1 of 4
1. Question
Solve for `x``log_a x+log_a 4=log_a (x+1)`- `x=` (1/3)
Hint
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Laws of Logarithms
$$\log_{\color{#9a00c7}{b}} {\color{#00880A}{x}\color{#e65021}{y}}=\log_{\color{#9a00c7}{b}} \color{#00880A}{x} + \log_{\color{#9a00c7}{b}} \color{#e65021}{y}$$Contract the left side$$\log_\color{#9a00c7}{a} \color{#00880A}{x}+\log_\color{#9a00c7}{a} \color{#e65021}{4}$$ `=` $$\log_a (x+1)$$ $$\log_\color{#9a00c7}{a} (\color{#00880A}{x})(\color{#e65021}{4})$$ `=` $$\log_a (x+1)$$ `log_b xy=log_b x+log_b y` $$\log_a 4x$$ `=` $$\log_a (x+1)$$ Since the bases of both sides are the same, the logarithm can be dropped$$\log_{a} \color{#00880A}{4x}$$ `=` $$\log_{a} \color{#00880A}{(x+1)}$$ $$\color{#00880A}{4x}$$ `=` $$\color{#00880A}{x+1}$$ `4x` `-x` `=` `x+1` `-x` Subtract `x` from both sides `3x` `=` `1` `3x``divide3` `=` `1``divide3` Divide both sides by `3` `x` `=` `1/3` `x=1/3` -
Question 2 of 4
2. Question
Solve for `x``log_2 x+log_2 (x+3)=2`- `x=` (1)
Hint
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Correct!
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Laws of Logarithms
$$\log_{\color{#9a00c7}{b}} \frac{\color{#00880A}{x}}{\color{#e65021}{y}}=\log_{\color{#9a00c7}{b}} \color{#00880A}{x}-\log_{\color{#9a00c7}{b}} \color{#e65021}{y}$$$$\log_b x^\color{#004ec4}{p}=\color{#004ec4}{p}\log_b x$$$$\log_{\color{#9a00c7}{b}} \color{#9a00c7}{b}=1$$Add a logarithmic term to the constant (third term)$$\log_{2} x+\log_{2} (x+3)$$ `=` $$2$$ $$\log_{2} x+\log_{2} (x+3)$$ `=` $$2\log_\color{#9a00c7}{2} \color{#9a00c7}{2}$$ `1=log_{2} 2` Remove the coefficient from the third term$$\log_{2} x+\log_{2} (x+3)$$ `=` $$2\log_{2} 2$$ $$\log_{2} x+\log_{2} (x+3)$$ `=` $$\log_{2} 2^\color{#004ec4}{2}$$ `log_b x^p=p log_b x` $$\log_{2} x+\log_{2} (x+3)$$ `=` $$\log_{2} 4$$ Contract the left side$$\log_\color{#9a00c7}{2} \color{#00880A}{x}+\log_\color{#9a00c7}{2} \color{#e65021}{(x+3)}$$ `=` $$\log_{2} 4$$ $$\log_\color{#9a00c7}{2} \color{#00880A}{x}\color{#e65021}{(x+3)}$$ `=` $$\log_{2} 4$$ `log_b xy=log_b x+log_b y` Since the bases of both sides are the same, the logarithm can be dropped$$\log_{2} \color{#00880A}{x(x+3)}$$ `=` $$\log_{2} \color{#00880A}{4}$$ $$\color{#00880A}{x(x+3)}$$ `=` $$\color{#00880A}{4}$$ $$x^2+3x$$ `=` $$4$$ Distribute `x^2+3x` `-4` `=` `4` `-4` Subtract `4` from both sides `(x+4)(x-1)` `=` `0` Factorize The possible values of `x` for this equation are `-4` and `1`Logarithms have to be positive. Therefore, `x=1``x=1` -
Question 3 of 4
3. Question
Solve for `x``log_a (x+2)-log_a (x-2)=log_a 5`- `x=` (3)
Hint
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Fantastic!
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Laws of Logarithms
$$\log_{\color{#9a00c7}{b}} \frac{\color{#00880A}{x}}{\color{#e65021}{y}}=\log_{\color{#9a00c7}{b}} \color{#00880A}{x}-\log_{\color{#9a00c7}{b}} \color{#e65021}{y}$$Contract the left side$$\log_\color{#9a00c7}{a} \color{#00880A}{(x+2)}-\log_\color{#9a00c7}{a} \color{#e65021}{(x-2)}$$ `=` $$\log_{a} 5$$ $$\log_\color{#9a00c7}{a} {\frac{\color{#00880A}{(x+2)}}{\color{#e65021}{(x-2)}}}$$ `=` $$\log_{a} 5$$ $$log_b \frac{x}{y}=log_b x-\log_b y$$ Since the bases of both sides are the same, the logarithm can be dropped$$\log_{a} \color{#00880A}{\frac{(x+2)}{(x-2)}}$$ `=` $$\log_{a} \color{#00880A}{5}$$ $$\color{#00880A}{\frac{(x+2)}{(x-2)}}$$ `=` $$\color{#00880A}{\frac{5}{1}}$$ $$5(x-2)$$ `=` $$x+2$$ Cross multiply $$5x-10$$ `=` $$x+2$$ Distribute `5x-10` `-x` `=` `x+2` `-x` Subtract `x` from both sides `4x-10` `=` `2` `4x-10` `+10` `=` `2` `+10` Add `10` to both sides `4x` `=` `12` `4x` `divide4` `=` `12` `divide4` Divide both sides by `4` `x` `=` `3` `x=3` -
Question 4 of 4
4. Question
Solve for `x``log_10 (x+7)-log_10 (x-2)=1`- `x=` (3)
Hint
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Nice Job!
Incorrect
Laws of Logarithms
$$\log_{\color{#9a00c7}{b}} \frac{\color{#00880A}{x}}{\color{#e65021}{y}}=\log_{\color{#9a00c7}{b}} \color{#00880A}{x}-\log_{\color{#9a00c7}{b}} \color{#e65021}{y}$$$$\log_{\color{#9a00c7}{b}} \color{#9a00c7}{b}=1$$Add a logarithmic term to the constant (third term)$$\log_{10} (x+7)-\log_{10} (x-2)$$ `=` $$1$$ $$\log_{10} (x+7)-\log_{10} (x-2)$$ `=` $$\log_\color{#9a00c7}{10} \color{#9a00c7}{10}$$ `1=log_{10} 10` Contract the left side$$\log_\color{#9a00c7}{10} \color{#00880A}{(x+7)}-\log_\color{#9a00c7}{10} \color{#e65021}{(x-2)}$$ `=` $$\log_{10} 10$$ $$\log_\color{#9a00c7}{10} {\frac{\color{#00880A}{(x+7)}}{\color{#e65021}{(x-2)}}}$$ `=` $$\log_{10} 10$$ $$log_b \frac{x}{y}=log_b x-\log_b y$$ Since the bases of both sides are the same, the logarithm can be dropped$$\log_{10} \color{#00880A}{\frac{(x+7)}{(x-2)}}$$ `=` $$\log_{10} \color{#00880A}{10}$$ $$\color{#00880A}{\frac{(x+7)}{(x-2)}}$$ `=` $$\color{#00880A}{\frac{10}{1}}$$ $$10(x-2)$$ `=` $$x+7$$ Cross multiply $$10x-20$$ `=` $$x+7$$ Distribute `10x-20` `-x` `=` `x+7` `-x` Subtract `x` from both sides `9x-20` `=` `7` `9x-20` `+20` `=` `7` `+20` Add `20` to both sides `9x` `=` `27` `9x` `divide9` `=` `27` `divide9` Divide both sides by `9` `x` `=` `3` `x=3`
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