SOLUTION: Working of Log 0.8621 to base 12 with detailed explanation

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Question 1210403: Working of Log 0.8621 to base 12 with detailed explanation
Found 3 solutions by MathLover1, ikleyn, Edwin McCravy:
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!

Use the change of base formula:

given: to base

in your case ,,

so, you have

≈ ≈


Answer by ikleyn(52921)   (Show Source): You can put this solution on YOUR website!
.
Working of Log 0.8621 to base 12 with detailed explanation
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        If I understand your writing correctly  (it is written in non-standard form for to be a proper  Math problem)
        you want  (or  " they "  want you)  calculate logarithm of  0.8621  in base  12.

        @MatLover1 interpreted it differently and gave you another answer.

        I will interpret it literally as I wrote it at the beginning of my post. 


Use the change of base formula    = .


Using your standard logarithmic table base 10 or a calculator

     = -0.064442355,   

         = 1.079181246.


THEREFORE, 

     =  =  = -0.059714117.


You may round your final answer in this way  

     = -0.05971  (approximately).    <<<---===  ANSWER

Solved.


----------------------------------


In standard  Math form,  your question should be


*************************************************************************

        Find the value of logarithm of  0.8621  base 12 using logarithms base 10.

*************************************************************************


Then it sounds unambiguously and does not create any distinct interpretations.



Answer by Edwin McCravy(20065)   (Show Source): You can put this solution on YOUR website!

Instead of using the memorized change of base formula, I prefer this:
 
a. Let the desired log equal x.
b. Use the definition of logarithm to write the equivalent exponential
   equation.
c. Take natural logs of both sides.
d. Move the exponent of x in front of the "ln" as a coefficient.
e. Solve for x. 













Edwin
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