$\log_{3}{(5x-2)}$ $-$ $2\log_{3}{\sqrt{3x+1}}$ $\,=\,$ $1-\log_{3}{4}$ is a logarithmic equation. It is developed in mathematics by taking number $3$ as base of the logarithms.
The square root of $3x+1$ can be eliminated from second term by the multiply factor $2$ as exponent of the $3x+1$. It can be done by using power rule of logarithms.
$\implies$ $\log_{3}{(5x-2)}$ $-$ $\log_{3}{{(\sqrt{3x+1})}^2}$ $\,=\,$ $1-\log_{3}{4}$
$\implies$ $\log_{3}{(5x-2)}$ $-$ $\log_{3}{(3x+1)}$ $\,=\,$ $1-\log_{3}{4}$
Make the logarithmic equation to have log terms one side and constant term in other side of the equation.
$\implies$ $\log_{3}{(5x-2)}$ $-$ $\log_{3}{(3x+1)}$ $+$ $\log_{3}{4}$ $\,=\,$ $1$
A negative sign between first two log terms represents subtraction of log terms. They can be combined by using quotient rule of logarithms.
$\implies$ $\log_{3}{\Bigg(\dfrac{5x-2}{3x+1}\Bigg)}$ $+$ $\log_{3}{4}$ $\,=\,$ $1$
A plus sign between the log terms expresses a summation of them. They can be merged as a logarithmic term by the product rule of logarithms.
$\implies$ $\log_{3}{\Bigg(\dfrac{4(5x-2)}{3x+1}\Bigg)}$ $\,=\,$ $1$
Write the logarithmic equation in exponential form equation by the mathematical relation between logarithms and exponents.
$\implies$ $\dfrac{4(5x-2)}{3x+1}$ $\,=\,$ $3^1$
$\implies$ $\dfrac{4(5x-2)}{3x+1}$ $\,=\,$ $3$
Use cross multiplication method to solve the algebraic equation and it evaluates the value of $x$.
$\implies$ $4(5x-2)$ $\,=\,$ $3(3x+1)$
$\implies$ $4 \times 5x-4 \times 2$ $\,=\,$ $3 \times 3x + 3 \times 1$
$\implies$ $20x-8$ $\,=\,$ $9x+3$
$\implies$ $20x-9x$ $\,=\,$ $3+8$
$\implies$ $11x$ $\,=\,$ $11$
$\implies$ $x$ $\,=\,$ $\dfrac{11}{11}$
$\implies$ $x$ $\,=\,$ $\require{cancel} \dfrac{\cancel{11}}{\cancel{11}}$
$\,\,\, \therefore \,\,\,\,\,\, x$ $\,=\,$ $1$
Thus, the log equation $\log_{3}{(5x-2)}$ $-$ $2\log_{3}{\sqrt{3x+1}}$ $\,=\,$ $1-\log_{3}{4}$ is solved by properties of logarithms in logarithmic mathematics.
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