Question 202664
Quick note: {{{log(7,(7))=1}}} and {{{log(5,(1))=0}}}. So {{{log(5,(log(7,(7))))=log(5,(1))=0}}}. In other words, {{{log(5,(log(7,(7))))=0}}}



So the expression {{{(1/3)log(20,(2x-1))=log(20,(11))-2*log(20,((2x-1)^(1/3)))+log(5,(log(7,(7))))}}}



simplifies to {{{(1/3)log(20,(2x-1))=log(20,(11))-2*log(20,((2x-1)^(1/3)))}}}



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{{{(1/3)log(20,(2x-1))=log(20,(11))-2*log(20,((2x-1)^(1/3)))}}} Start with the given expression.



{{{(1/3)log(20,(2x-1))=log(20,(11))-2*(1/3)log(20,(2x-1))}}} Rewrite the last log using the identity  {{{log(b,(x^y))=y*log(b,(x))}}}



{{{(1/3)log(20,(2x-1))=log(20,(11))-(2/3)log(20,(2x-1))}}} Multiply



{{{(1/3)log(20,(2x-1))-log(20,(11))=-(2/3)log(20,(2x-1))}}} Subtract {{{log(20,(11))}}} from both sides.



{{{-log(20,(11))=-(2/3)log(20,(2x-1))-(1/3)log(20,(2x-1))}}} Subtract {{{(1/3)log(20,(2x-1))}}} from both sides.



{{{-log(20,(11))=(-2/3-1/3)log(20,(2x-1))}}} Factor out the GCF {{{log(20,(2x-1))}}}



{{{-log(20,(11))=(-3/3)log(20,(2x-1))}}} Combine like terms.



{{{-log(20,(11))=-log(20,(2x-1))}}} Reduce



{{{log(20,(11))=log(20,(2x-1))}}} Multiply both sides by -1



{{{11=2x-1}}} Since the bases of the logs are equal, this means that the arguments are equal.



{{{11+1=2x}}} Add 1 to both sides.



{{{12=2x}}} Combine like terms.



{{{12/2=x}}} Divide both sides by 2 to isolate "x".



{{{6=x}}} Reduce



{{{x=6}}} Rearrange the equation.



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Answer:



So the solution is {{{x=6}}}