What is the solution of Log of 0.00001 to the base 0.01 without using logarithm table and calculator?

1 Answer
Mar 8, 2018

"The answer is:" \qquad \qquad \qquad \quad log_{0.01} 0.00001 \ = \ 5/2.

Explanation:

"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.