Difference between revisions of "004 Sample Final A, Problem 19"
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Kayla Murray (talk | contribs) (Created page with "<span class="exam"> Solve for ''x'': <math>\log_6 \frac{1}{36} = x</math>") |
Kayla Murray (talk | contribs) |
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<span class="exam"> Solve for ''x'': <math>\log_6 \frac{1}{36} = x</math> | <span class="exam"> Solve for ''x'': <math>\log_6 \frac{1}{36} = x</math> | ||
+ | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | !Foundations | ||
+ | |- | ||
+ | |How do we remove logs from an equation? | ||
+ | |- | ||
+ | |Answer: | ||
+ | |- | ||
+ | |The definition of the 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: | ||
+ | |- | ||
+ | |By the definition of the logarithm, <math>\log_6 \frac{1}{36} = x</math> means <math>6^x=\frac{1}{36}</math>. | ||
+ | |} | ||
+ | |||
+ | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Step 2: | ||
+ | |- | ||
+ | |Now, we can solve for <math>x</math>. Since <math>6^x=\frac{1}{36}</math>, | ||
+ | |- | ||
+ | |we must have <math>x=-2</math>. | ||
+ | |} | ||
+ | |||
+ | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Final Answer: | ||
+ | |- | ||
+ | | <math>x=-2</math> | ||
+ | |} | ||
+ | |||
+ | [[004 Sample Final A|<u>'''Return to Sample Exam</u>''']] |
Latest revision as of 10:17, 29 April 2015
Solve for x:
Foundations |
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How do we remove logs from an equation? |
Answer: |
The definition of the logarithm tells us that if , then . |
Solution:
Step 1: |
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By the definition of the logarithm, means . |
Step 2: |
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Now, we can solve for . Since , |
we must have . |
Final Answer: |
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