Question 140699
I'm assuming that the problem looks like this: {{{log(64,(8))=x}}}???



{{{log(64,(8))=x}}} Start with the given equation



{{{log(10,(8))/log(10,(64))=x}}} Rewrite the left side by using the change of base formula. Remember the change of base formula is {{{log(b,(x))=log(10,(x))/log(10,(b))}}}



{{{log(10,(8))/log(10,(8^2))=x}}} Rewrite 64 as {{{8^2}}}



{{{log(10,(8))/(2*log(10,(8)))=x}}} Rewrite the denominator using the identity  {{{log(b,(x^y))=y*log(b,(x))}}}



{{{cross(log(10,(8)))/(2cross(log(10,(8))))=x}}} Divide and cancel like terms



{{{1/2=x}}} Simplify



So the answer is {{{x=1/2}}}



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{{{log(5,(1/5))=x}}} Start with the given equation




{{{log(10,(1/5))/log(10,(5))=x}}} Rewrite the left side by using the change of base formula. 





{{{log(10,(1/5))/log(10,((1/5)^(-1)))=x}}} Rewrite 5 as {{{(1/5)^(-1)}}}. Notice how {{{5=(1/5)^(-1)}}}



{{{log(10,(1/5))/(-1*log(10,(1/5)))=x}}} Rewrite the denominator using the identity  {{{log(b,(x^y))=y*log(b,(x))}}}



{{{cross(log(10,(1/5)))/(-1*cross(log(10,(1/5))))=x}}} Cancel like terms



{{{1/(-1)=x}}} Simplify


{{{x=-1}}} Reduce



So the answer is {{{x=-1}}} 



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{{{log(7,(sqrt(7)))=x}}} Start with the given equation




{{{log(7,(7^(1/2)))=x}}} Rewrite {{{sqrt(7)}}} as {{{7^(1/2)}}}


{{{(1/2)*log(7,(7))=x}}} Rewrite the expression using the identity  {{{log(b,(x^y))=y*log(b,(x))}}}



{{{(1/2)*1=x}}} Take the log base 7 of 7 to get 1. In other words, {{{log(7,(7))=1}}}



{{{x=1/2}}} Multiply




So the answer is {{{x=1/2}}}