Question 198808
a)


Since you cannot take the log of 0 or a negative value, this means that {{{log(10,(x))>0}}} and {{{x>0}}} (for the outer and inner logs)



If we solve {{{log(10,(x))>0}}}, we get {{{x>1}}}. So combining the inequalities {{{x>0}}} and {{{x>1}}} (ie perform a set union), we get {{{x>1}}}.


So the domain of {{{f(x)=log(5,(log(10,(x))))}}} is {{{x>1}}}



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b)


{{{f(x)=log(5,(log(10,(x))))}}} Start with the given function.



{{{y=log(5,(log(10,(x))))}}} Replace f(x) with "y"



{{{x=log(5,(log(10,(y))))}}} Switch each x and y



{{{5^x=log(10,(y))}}} Rewrite the equation using the property: {{{log(b,(x))=y}}} ====> {{{b^y=x}}}



{{{10^(5^x)=y}}} Rewrite the equation again using the property: {{{log(b,(x))=y}}} ====> {{{b^y=x}}}



So the inverse function is *[Tex \LARGE f^{-1}(x)=10^^{5^{x}}]