Thursday, May 9, 2013

how do you do log7(2-x)=log7(5x)


we use the inverse property of logs/exponents: 

%0D%0A+log7%282-x%29=log7%285x%29%0D%0A


%0D%0A7%5E+log7%282-x%29=7%5Elog7%285x%29%0D%0A


and the inputs are recovered:
%0D%0A2-x=5x%0D%0A



x=1%2F3%0D%0A
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