Difference between revisions of "008A Sample Final A, Question 16"
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| − | !Foundations | + | !Foundations: |
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|1) How do we combine the two logs? | |1) How do we combine the two logs? | ||
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| − | ! Step 1: | + | !Step 1: |
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|Using one of the properties of logarithms the, left hand side is equal to <math> \log_6( (x + 2)(x - 3)</math> | |Using one of the properties of logarithms the, left hand side is equal to <math> \log_6( (x + 2)(x - 3)</math> | ||
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| − | ! Step 2: | + | !Step 2: |
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|By the definition of logarithms <math> \log_6( (x + 2)(x - 3) = 1</math> means <math> 6 = (x + 2)(x - 3)</math> | |By the definition of logarithms <math> \log_6( (x + 2)(x - 3) = 1</math> means <math> 6 = (x + 2)(x - 3)</math> | ||
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| − | ! Step 3: | + | !Step 3: |
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|Now we do some arithmetic to solve for x. <math> 0 = (x + 2)(x - 3) - 6 = x^2 - x - 12 = (x - 4)(x + 3) </math>. So there are two possible answers. | |Now we do some arithmetic to solve for x. <math> 0 = (x + 2)(x - 3) - 6 = x^2 - x - 12 = (x - 4)(x + 3) </math>. So there are two possible answers. | ||
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| − | ! Step 4: | + | !Step 4: |
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|We have to make sure the answers make sense in the context of the problem. Since the domain of the log function is <math> (0, \infty)</math> , -3 is removed as a potential answer. | |We have to make sure the answers make sense in the context of the problem. Since the domain of the log function is <math> (0, \infty)</math> , -3 is removed as a potential answer. | ||
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| − | ! Final Answer: | + | !Final Answer: |
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| x = 4. | | x = 4. | ||
Latest revision as of 23:02, 25 May 2015
Question: Solve. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \log_6(x+2)+\log_6(x-3) = 1 }
| Foundations: |
|---|
| 1) How do we combine the two logs? |
| 2) How do we remove the logs? |
| Answer: |
| 1) One of the rules of logarithms says that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \log(x) + \log(y) = \log(xy)} |
| 2) The definition of logarithm tells us that if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \log_6(x) = y } , then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 6^y = x } |
Solution:
| Step 1: |
|---|
| Using one of the properties of logarithms the, left hand side is equal to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \log_6( (x + 2)(x - 3)} |
| Step 2: |
|---|
| By the definition of logarithms Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \log_6( (x + 2)(x - 3) = 1} means Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 6 = (x + 2)(x - 3)} |
| Step 3: |
|---|
| Now we do some arithmetic to solve for x. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 = (x + 2)(x - 3) - 6 = x^2 - x - 12 = (x - 4)(x + 3) } . So there are two possible answers. |
| Step 4: |
|---|
| We have to make sure the answers make sense in the context of the problem. Since the domain of the log function is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (0, \infty)} , -3 is removed as a potential answer. |
| Final Answer: |
|---|
| x = 4. |