$${\mathtt{100}} = {\mathtt{10}}{\mathtt{\,\times\,}}{log}_{10}\left({\frac{{\mathtt{z}}}{{{\mathtt{10}}}^{-{\mathtt{12}}}}}\right)$$ I know the answer is $${{\mathtt{10}}}^{-{\mathtt{2}}}$$ but how do I get to it? Thank you.
+118723 $$\begin{array}{rll}
100&=&10\times log_{10}\left(\frac{z}{10^{-12}}\right)\\\\
10&=&log_{10}\left(\frac{z}{10^{-12}}\right)\\\\
10^{10}&=&10^{log_{10}\left(\frac{z}{10^{-12}}\right)}\\\\
10^{10}&=&\frac{z}{10^{-12}}\\\\
10^{10}\times 10^{-12}&=&z\\\\
z&=&10^{10-12}\\\\
z&=&10^{-2}
\end{array}$$
+23254 100 = 10 · log( z / 10^-12)
100 = 10 · log( 10^12 · z )
100 = 10 · [ log( 10^12) + log( z ) ] (When numbers are multiplied, the logs are added.)
100 = 10 · [ 12 · log( 10 ) + log( z ) ] (Exponents come out as multipliers.)
100 = 10 · [ 12 + log( z ) ] (log( 10 ) = 1 ---> 12 · log( 10 ) = 12 · 1 = 12 )
100 = 120 + 10 · log( z ) (Distribute the 10.)
-20 = 10 · log( z ) (Subtract 120 from both sides.)
-2 = log( z ) (Divide both sides by 10.)
---> z = 10^-2 (Switch from log form to exponential form.)
+118723 $$\begin{array}{rll}
100&=&10\times log_{10}\left(\frac{z}{10^{-12}}\right)\\\\
10&=&log_{10}\left(\frac{z}{10^{-12}}\right)\\\\
10^{10}&=&10^{log_{10}\left(\frac{z}{10^{-12}}\right)}\\\\
10^{10}&=&\frac{z}{10^{-12}}\\\\
10^{10}\times 10^{-12}&=&z\\\\
z&=&10^{10-12}\\\\
z&=&10^{-2}
\end{array}$$
