$\log_{3}{(5x-2)}$ $-$ $2\log_{3}{\sqrt{3x+1}}$ $\,=\,$ $1-\log_{3}{4}$ 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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