Difference between revisions of "005 Sample Final A, Question 8"

Latest revision as of 20:25, 21 May 2015

Question Solve the following equation, $3^{2x}+3^{x}-2=0$

Foundations
1) What substitution can we make to simplify the problem?
Answer:
1) Substitute $y=3^{x}$ to change the original equation into $y^{2}+y-2=0$

Step 1:
Start by rewriting $3^{2x}=\left(3^{x}\right)^{2}$ and make the substitution $y=3^{x}$

Step 2:
After substitution we get $y^{2}+y-2=(y+2)(y-1)$

Step 3:
Now we have to find the zeros of $3^{x}+2=0$ and $3^{x}-1=0$ . We do this by first isolating $3^{x}$ in both equations.
So $3^{x}=-2$ and $3^{x}=1$

Step 4:
We observe that $3^{x}=-2$ has no solutions. We can solve $3^{x}=1$ by taking $log_{3}$ of both sides.
This gives $\log _{3}\left(3^{x}\right)=x=\log _{3}(1)=0$

Final Answer:
x = 0

@@ Line 3: / Line 3: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Final Answers
+!Foundations
 |-
-|a) False. Nothing in the definition of a geometric sequence requires the common ratio to be always positive. For example, <math>a_n = (-a)^n</math>
+|1) What substitution can we make to simplify the problem?
 |-
-|b) False. Linear systems only have a solution if the lines intersect. So y = x and y = x + 1 will never intersect because they are parallel.
+|Answer:
 |-
-|c) False. <math>y = x^2</math> does not have an inverse.
+|1) Substitute <math>y = 3^x</math> to change the original equation into <math>y^2 + y - 2 = 0</math>
+|}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+! Step 1:
 |-
-|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
+| Start by rewriting <math>3^{2x} = \left(3^x\right)^2</math> and make the substitution <math>y = 3^x</math>
+|}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+! Step 2:
 |-
-|e) True.
+| After substitution we get <math>y^2 + y - 2 = (y + 2)(y - 1)</math>
+|}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+! Step 3:
 |-
-|f) False.
+| Now we have to find the zeros of <math>3^x + 2 = 0</math> and <math>3^x - 1 = 0</math>. We do this by first isolating <math>3^x</math> in both equations.
+|-
+|So <math>3^x = -2</math> and <math>3^x = 1</math>
+|}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+! Step 4:
+|-
+| We observe that <math>3^x = -2</math> has no solutions. We can solve <math>3^x = 1</math> by taking <math>log_3</math> of both sides.
+|-
+|This gives<math>\log_3\left(3^x\right) = x = \log_3(1) = 0</math>
+|}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+! Final Answer:
+|-
+| x = 0
 |}

Difference between revisions of "005 Sample Final A, Question 8"

Latest revision as of 20:25, 21 May 2015

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