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Question about logarithm

  • Thread starter Coco12
  • Start date
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1. The problem statement, all variables and given/known data

6^log x=1/36

2. Relevant equations
Y=logcx

3. The attempt at a solution
How do u solve this? I know the 1/36 is the exponent. Usually the logs that u normally do is not in an exponent like this. The answer is 0.01. How did they get that?
 

Dick

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Homework Helper
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1. The problem statement, all variables and given/known data

6^log x=1/36

2. Relevant equations
Y=logcx

3. The attempt at a solution
How do u solve this? I know the 1/36 is the exponent. Usually the logs that u normally do is not in an exponent like this. The answer is 0.01. How did they get that?


Take the log of both sides. Use rules of logs and solve for log(x). If you want to take a shortcut, 1/36=6^(-2), right?
 
272
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I thought the ans was -2 since 6^-2 = 1/36

But the ans is 0.01. So that means it is not...
 

Dick

Science Advisor
Homework Helper
26,252
615
I thought the ans was -2 since 6^-2 = 1/36

But the ans is 0.01. So that means it is not...
log(x)=(-2). x isn't -2. You are supposed to solve for x. What's the base of your logarithms?
 
272
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Is the base 10?
 

Dick

Science Advisor
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Is the base 10?
It should be stated in the problems or your book what the base is. Suppose it is 10. Then if log(x)=(-2), what is x?
 

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