Answer
$-0.1303$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the Laws of Logarithms to find an equivalent expression for the given expression, $
\log_{10} \dfrac{20}{27}
,$ which uses the given logarithmic values, $\log_{10} 2=
0.3010
$ and/or $\log_{10} 3=
0.4771
.$
$\bf{\text{Solution Details:}}$
The given expression can be expressed as
\begin{array}{l}\require{cancel}
\log_{10} \dfrac{10\cdot2}{3^3}
.\end{array}
Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
\log_{10} (10\cdot2)-\log_{10}3^3
.\end{array}
Using the Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the given expression is equivalent
\begin{array}{l}\require{cancel}
\log_{10} 10+\log_{10} 2-\log_{10}3^3
.\end{array}
Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
\log_{10} 10+\log_{10} 2-3\log_{10}3
.\end{array}
Since $\log_b b=1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
1+\log_{10} 2-3\log_{10}3
.\end{array}
Substituting the known values of the logarithmic expressions results to
\begin{array}{l}\require{cancel}
1+0.3010-3(0.4771)
\\\\=
1+0.3010-1.4313
\\\\=
-0.1303
.\end{array}
