$\begin{array} { l }2\log_{ 10 }({ x })-\log_{ 10 }({ 7x })-1=0,& x \in \langle0, +\infty\rangle\end{array}$
Use $\log_{ a }({ x \times y })=\log_{ a }({ x })+\log_{ a }({ y })$ to expand the expression$2\log_{ 10 }({ x })-\left( \log_{ 10 }({ 7 })+\log_{ 10 }({ x }) \right)-1=0$
When there is a $-$ in front of an expression in parentheses, change the sign of each term of the expression and remove the parentheses$2\log_{ 10 }({ x })-\log_{ 10 }({ 7 })-\log_{ 10 }({ x })-1=0$
Collect like terms$\log_{ 10 }({ x })-\log_{ 10 }({ 7 })-1=0$
Move the constants to the right-hand side and change their signs$\log_{ 10 }({ x })=\log_{ 10 }({ 7 })+1$
$1$ can be expressed as a logarithm with the same base and argument$\log_{ 10 }({ x })=\log_{ 10 }({ 7 })+\log_{ 10 }({ 10 })$
Calculate the sum of logarithms$\log_{ 10 }({ x })=\log_{ 10 }({ 70 })$
Since the bases of the logarithms are the same, set the arguments equal$\begin{array} { l }x=70,& x \in \langle0, +\infty\rangle\end{array}$
Check if the solution is in the defined range$x=70$