Difference between revisions of "004 Sample Final A, Problem 15"
Jump to navigation
Jump to search
Kayla Murray (talk | contribs) |
Kayla Murray (talk | contribs) |
||
| (2 intermediate revisions by the same user not shown) | |||
| Line 3: | Line 3: | ||
!Foundations | !Foundations | ||
|- | |- | ||
| − | | | + | |1) How can we combine the two logs? |
|- | |- | ||
| − | | | + | |2) How do we remove logs from an equation? |
|- | |- | ||
|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 10:36, 29 April 2015
Solve.
| 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 |
| 2) The definition of the logarithm tells us that if , then . |
Solution:
| Step 1: |
|---|
| Using a rule of logarithms, the equation becomes . |
| Step 2: |
|---|
| By the definition of the logarithm, |
| means |
| Step 3: |
|---|
| Now, we can solve for . We have . |
| So, there are two possible answers, which are or . |
| 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
, -9 is removed as a potential answer. The answer is . |
| Final Answer: |
|---|