Question 532404
How do I plug a LOG problem that is not base 10?  The question is Log7(1/49).  
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One way is to use the base change formula, it always works.
{{{log(7,1/49) = log(1/49)/log(7)}}}
= -2
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Another is to recognize that {{{1/49 = 7^-2}}}
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Not sure what you mean my "plug."