Question 113827
Remember if we have {{{log(b,(x))=y}}}, we can write it as {{{b^y=x}}}



a)

{{{log(1/2,(4))=y}}} Start with the given equation. Just set the original expression equal to y. Note: the entire {{{1/2}}} is the base of the log.



{{{(1/2)^y=4}}} Rewrite the original expression using the property I listed above




{{{(2^(-1))^y=4}}} Rewrite {{{1/2}}} as {{{2^(-1)}}}



{{{2^(-y)=4}}} Multiply the exponents 




{{{2^(-y)=2^2}}} Rewrite {{{4}}} as {{{2^2}}} 



Since the bases are equal, the exponents are equal. So {{{-y=2}}} which means {{{y=-2}}}


So {{{log(1/2,(4))=-2}}}



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b)

{{{log(8,(64))=y}}} Start with the given equation. Just set the original expression equal to y



{{{8^y=64}}} Rewrite the original expression using the property: {{{log(b,(x))=y}}} <===> {{{b^y=x}}}



{{{8^y=8^2}}} Rewrite {{{64}}} as {{{8^2}}}



Since the bases are equal, the exponents are equal. So {{{y=2}}}


So {{{log(8,(64))=2}}} 



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c)


{{{log(32,(8))=y}}} Start with the given equation. Just set the original expression equal to y



{{{32^y=8}}} Rewrite the original expression using the property: {{{log(b,(x))=y}}} <===> {{{b^y=x}}}



{{{(2^5)^y=2^3}}} Rewrite {{{8}}} as {{{2^3}}} and {{{32}}} as {{{2^5}}}



{{{2^(5y)=2^3}}} Multiply the exponents 


Since the bases are equal, the exponents are equal. So {{{5y=3}}} which means {{{y=3/5}}}


So {{{log(32,(8))=3/5}}} 


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d)


{{{log(3,(81))=y}}} Start with the given equation. Just set the original expression equal to y



{{{3^y=81}}} Rewrite the original expression using the property: {{{log(b,(x))=y}}} <===> {{{b^y=x}}}



{{{3^y=3^4}}} Rewrite {{{81}}} as {{{3^4}}}


Since the bases are equal, the exponents are equal. So {{{y=4}}}


So {{{log(3,(81))=4}}}


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e)


{{{pi^(log(pi,(2pi)))}}} Start with the given expression



{{{pi^(log(pi,(2))+log(pi,(pi)))}}} Break up the logarithm using the identity {{{log(b,(x*y))=log(b,(x))+log(b,(y))}}}. Think of {{{2pi}}} as 2 times {{{pi}}}.



{{{pi^(log(pi,(2)))*pi^(log(pi,(pi)))}}} Break up the exponent using the identity {{{b^(x+y)=b^x*b^y}}}



{{{pi^(log(pi,(2)))*pi^1}}} Evaluate {{{log(pi,(pi))}}} to get 1


Since we cannot simplify the expression {{{log(pi,(2))}}} any further, we cannot simplify the entire expression any further.