008A Sample Final A, Question 16 - Revision history

MathAdmin at 07:02, 26 May 2015

2015-05-26T07:02:40Z

MathAdmin at 23:13, 23 May 2015

2015-05-23T23:13:36Z

MathAdmin: Created page with "'''Question: ''' Solve. $\log_6(x+2)+\log_6(x-3) = 1$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations |- |1) How do we combine th..."

2015-04-28T21:14:13Z

Created page with "'''Question: ''' Solve. <math> \log_6(x+2)+\log_6(x-3) = 1 </math> {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations |- |1) How do we combine th..."

New page

'''Question: ''' Solve. <math> \log_6(x+2)+\log_6(x-3) = 1 </math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!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 <math> \log(x) + \log(y) = \log(xy)</math>
|-
|2) The definition of logarithm tells us that if <math> \log_6(x) = y </math>, then <math> 6^y = x </math>
|}

Solution:

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 1:
|-
|Using a rule of logarithms the left hand side is equal to <math> \log_6( (x + 2)(x - 3)</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 2:
|-
|By the definition of logarithms <math> \log_6( (x + 2)(x - 3) = 1</math> means <math> 6 = (x + 2)(x - 3)</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 3:
|-
|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.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! 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>  ,   -3 is removed as a potential answer.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answer:
|-
| x = 4.
|}

[[008A Sample Final A|<u>'''Return to Sample Exam</u>''']]

@@ Line 2: / Line 2: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Foundations
+!Foundations: &nbsp;
 |-
 |1) How do we combine the two logs?
@@ Line 18: / Line 18: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 1:
+!Step 1: &nbsp;
 |-
 |Using one of the properties of logarithms the, left hand side is equal to <math> \log_6( (x + 2)(x - 3)</math>
@@ Line 24: / Line 24: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 2:
+!Step 2: &nbsp;
 |-
 |By the definition of logarithms <math> \log_6( (x + 2)(x - 3) = 1</math> means <math> 6 = (x + 2)(x - 3)</math>
@@ Line 30: / Line 30: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 3:
+!Step 3: &nbsp;
 |-
 |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.
@@ Line 36: / Line 36: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 4:
+!Step 4: &nbsp;
 |-
 |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>&nbsp; , &nbsp; -3 is removed as a potential answer.
@@ Line 42: / Line 42: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Final Answer:
+!Final Answer: &nbsp;
 |-
 | x = 4.

008A Sample Final A, Question 16 - Revision history

MathAdmin at 07:02, 26 May 2015

MathAdmin at 23:13, 23 May 2015

MathAdmin: Created page with "'''Question: ''' Solve. \log_6(x+2)+\log_6(x-3) = 1 {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations |- |1) How do we combine th..."

MathAdmin: Created page with "'''Question: ''' Solve. $\log_6(x+2)+\log_6(x-3) = 1$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations |- |1) How do we combine th..."