8A F11 Q16 - Revision history

Matt Lee at 07:07, 14 April 2015

2015-04-14T07:07:55Z

Matt Lee at 07:07, 14 April 2015

2015-04-14T07:07:35Z

Matt Lee at 07:07, 14 April 2015

2015-04-14T07:07:17Z

Matt Lee at 23:01, 6 April 2015

2015-04-06T23:01:01Z

Matt Lee: 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..."

2015-03-29T18:15:38Z

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.
|}

@@ Line 38: / Line 38: @@
 ! 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>,&nbsp; &nbsp; -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>&nbsp; , &nbsp; -3 is removed as a potential answer.
 |}

@@ Line 38: / Line 38: @@
 ! 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.
+|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.
 |}

← Older revision		Revision as of 23:01, 6 April 2015
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	\| x = 4.		\| x = 4.
	\|}		\|}
		+
		+	[[8AF11Final\|<u>'''Return to Sample Exam</u>''']]

8A F11 Q16 - Revision history

Matt Lee at 07:07, 14 April 2015

Matt Lee at 07:07, 14 April 2015

Matt Lee at 07:07, 14 April 2015

Matt Lee at 23:01, 6 April 2015

Matt Lee: 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..."

Matt Lee: 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..."