Difference between revisions of "004 Sample Final A, Problem 15"
Jump to navigation
Jump to search
Kayla Murray (talk | contribs) |
Kayla Murray (talk | contribs) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 9: | Line 9: | ||
|Answer: | |Answer: | ||
|- | |- | ||
| − | | | + | |1) One of the rules of logarithms states that <math>\log(x)+\log(y)=\log(xy) </math> |
|- | |- | ||
| − | | | + | |2) The definition of the logarithm tells us that if <math>\log(x)=y</math>, then <math>10^y=x</math>. |
|} | |} | ||
| Line 19: | Line 19: | ||
! Step 1: | ! Step 1: | ||
|- | |- | ||
| − | | | + | |Using a rule of logarithms, the equation becomes <math>\log((x+8)(x-1))=1</math>. |
|} | |} | ||
| Line 25: | Line 25: | ||
! Step 2: | ! Step 2: | ||
|- | |- | ||
| − | | | + | |By the definition of the logarithm, <math>\log((x+8)(x-1))=1</math> |
| + | |- | ||
| + | |means <math>10=(x+8)(x-1)</math> | ||
|} | |} | ||
| Line 31: | Line 33: | ||
! Step 3: | ! Step 3: | ||
|- | |- | ||
| − | | | + | |Now, we can solve for <math>x</math>. We have <math>0=(x+8)(x-1)-10=x^2+7x-18=(x+9)(x-2)</math>. |
| + | |- | ||
| + | |So, there are two possible answers, which are <math>x=-9</math> or <math>x=2</math>. | ||
|} | |} | ||
| Line 37: | Line 41: | ||
! Step 4: | ! 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 |
| + | <math> (0, \infty)</math>, -9 is removed as a potential answer. The answer is <math>x=2</math>. | ||
|} | |} | ||
| Line 43: | Line 48: | ||
! Final Answer: | ! Final Answer: | ||
|- | |- | ||
| − | | | + | | <math>x=2</math> |
|} | |} | ||
[[004 Sample Final A|<u>'''Return to Sample Exam</u>''']] | [[004 Sample Final A|<u>'''Return to Sample Exam</u>''']] | ||
Latest revision as of 09:36, 29 April 2015
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(x + 8) + \log(x - 1) = 1}
| Foundations |
|---|
| 1) How can we combine the two logs? |
| 2) How do we remove logs from an equation? |
| Answer: |
| 1) One of the rules of logarithms states 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 the 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(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 10^y=x} . |
Solution:
| Step 1: |
|---|
| Using a rule of logarithms, the equation becomes 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+8)(x-1))=1} . |
| Step 2: |
|---|
| By the definition of the logarithm, 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+8)(x-1))=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 10=(x+8)(x-1)} |
| Step 3: |
|---|
| Now, we can solve for 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 x} . We have 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+8)(x-1)-10=x^2+7x-18=(x+9)(x-2)} . |
| So, there are two possible answers, which are 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 x=-9} or 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 x=2} . |
| 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)} , -9 is removed as a potential answer. The answer 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 x=2} . |
| Final Answer: |
|---|
| 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 x=2} |