# "Nice question !! We can proceed as follows, using some Rules" #
# "for Logarithms, and some Rules for Exponents." #
# "One way to start is:" #
# "Let:" \qquad \qquad \qquad \qquad \qquad \qquad \qquad log_{0.01} 0.00001 \ = \ x. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ (I) #
# \qquad color{blue}{ "now rewrite this as an exponential equation using" } #
# \qquad \qquad \quad \ \ color{blue}{ "the fundamental property of logarithms:" } #
# \qquad \qquad \qquad \qquad \qquad \qquad color{blue}{ log_{b} a \ = \ x \quad iff \quad b^x = a. } #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ log_{0.01} 0.00001 \ = \ x #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ log_{0.01} 0.00001 \ = \ color{red}{x}#
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 0.01^color{red}{x} \ = \ 0.00001 #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ ( 10^{-2} )^x \ = \ 10^{-5} #
# \qquad color{blue}{ "now use Product Rule for Exponents:" \qquad ( a^p )^q \ = \ a^{ p cdot q }. #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ 10^{ -2 x } \ = \ 10^{-5} #
# \qquad color{blue}{ "now equate Exponents, because the base is the same on" } #
# \qquad color{blue}{ "both sides ---" \ \ 10. \ \ "[Note: cannot do this if the base" } #
# \qquad color{blue}{ "is 0 or 1; so we're Ok here with base 10.]" } #
# \qquad \qquad :. \qquad \qquad \qquad \qquad \quad \ -2 x \ = \ -5 #
# \qquad color{blue}{ "solve:" } #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ x \ = \ { -5 }/{ -2 } #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ x \ = \ 5/2. #
# "Now looking back to line (I), we see" \ \ x \ \ "is the original quantity" #
# "we wanted to find ! So:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad log_{0.01} 0.00001 \ = \ 5/2. #
