What is the solution to 6^log x = 1/36?

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To solve the equation 6^log x = 1/36, it is essential to recognize that 1/36 can be rewritten as 6^(-2). Taking the logarithm of both sides leads to log(x) = -2, which implies that x must be calculated as 10^(-2) if the logarithm base is 10. This results in x equaling 0.01. The confusion arises from interpreting the logarithm's base, which should be specified in the problem. Ultimately, the correct solution is x = 0.01.
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Homework Statement



6^log x=1/36

2. Homework Equations
Y=logcx

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?
 
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Coco12 said:

Homework Statement



6^log x=1/36

2. Homework Equations
Y=logcx

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?
 
I thought the ans was -2 since 6^-2 = 1/36

But the ans is 0.01. So that means it is not...
 
Coco12 said:
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?
 
Is the base 10?
 
Coco12 said:
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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