Difference between revisions of "008A Sample Final A, Question 1"

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! Step 3:
 
! Step 3:
 
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|Starting with <math>x = \log_3(y + 3) - 1</math>, we start by adding 1 to both sides to get
+
|From <math>x = \log_3(y + 3) - 1</math>, we add 1 to both sides to get
 
|-
 
|-
 
|<math>x + 1 = \log_3(y + 3).</math> Now we will use the relation in Foundations 2) to swap the log for an exponential to get
 
|<math>x + 1 = \log_3(y + 3).</math> Now we will use the relation in Foundations 2) to swap the log for an exponential to get
 
|-
 
|-
|<math>y + 3 = 3^{x+1}</math>. All we have to do is subtract 3 from both sides to yield the final answer
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|<math>y + 3 = 3^{x+1}</math>.
 
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Revision as of 21:55, 22 May 2015

Question: Find for


Foundations
1) How would you find the inverse for a simpler function like ?
2) How do you remove the in the following equation:
Answers:
1) you would replace f(x) by y, switch x and y, and finally solve for y.
2) By the definition of when we write the equation we mean y is the number such that


Solution:

Step 1:
We start by replacing f(x) with y.
This leaves us with
Step 2:
Now we swap x and y to get
In the next step we will solve for y.
Step 3:
From , we add 1 to both sides to get
Now we will use the relation in Foundations 2) to swap the log for an exponential to get
.
Step 4:
After subtracting 3 from both sides we get . Replacing y with we arrive at the final answer that
Final Answer:

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