The common logarithm of $7$ is added to the logarithm of $3x$ minus $2$ to base $10$ on left-hand side of the equation. Similarly, the common logarithm of $x$ plus $3$ is added to the number one on right hand side of the equation. It is given that the two mathematical expressions are equal.

$\log{7}$ $+$ $\log{(3x-2)}$ $\,=\,$ $\log{(x+3)}$ $+$ $1$

Let us learn how to solve the logarithmic equation to find the value of $x$.

Two logarithmic terms are added on left-hand side of the equation. The sum of them can be calculated by using the product rule of the logarithms.

$\implies$ $\log{\big(7 \times (3x-2)\big)}$ $\,=\,$ $\log{(x+3)}$ $+$ $1$

$\implies$ $\log{\big(7(3x-2)\big)}$ $\,=\,$ $\log{(x+3)}$ $+$ $1$

On right hand side of the equation, the common logarithm of $x$ plus $3$ is added to the number $1$ but they cannot be added due to their mathematical expression. However, the number $1$ can be expressed in common logarithm form.

$\implies$ $\log{\big(7(3x-2)\big)}$ $\,=\,$ $\log{(x+3)}$ $+$ $\log{10}$

Now, add the logarithmic terms as per the product rule of the logarithms.

$\implies$ $\log{\big(7(3x-2)\big)}$ $\,=\,$ $\log{\big((x+3) \times 10\big)}$

$\implies$ $\log{\big(7(3x-2)\big)}$ $\,=\,$ $\log{\big(10 \times (x+3)\big)}$

$\implies$ $\log{\big(7(3x-2)\big)}$ $\,=\,$ $\log{\big(10(x+3)\big)}$

The logarithmic equation expresses that the common logarithm of a function is equal to the common logarithm of another function. Therefore, the functions inside the logarithm should be equal.

$\implies$ $7(3x-2)$ $\,=\,$ $10(x+3)$

Now, simplify this algebraic equation to obtain the value of the $x$.

$\implies$ $7 \times (3x-2)$ $\,=\,$ $10 \times (x+3)$

$\implies$ $21x-14$ $\,=\,$ $10x+30$

$\implies$ $21x-10x$ $\,=\,$ $14+30$

$\implies$ $11x \,=\, 44$

$\implies$ $x \,=\, \dfrac{44}{11}$

$\implies$ $x \,=\, \dfrac{\cancel{44}}{\cancel{11}}$

$\,\,\,\therefore \,\,\,\,\,\,$ $x \,=\, 4$

Therefore, the value of the $x$ is $4$ and it is the required solution for this logarithm problem.

It is time to validate the zero or root of the given logarithmic equation.

$\log{7}$ $+$ $\log{(3x-2)}$ $=$ $\log{(x+3)}$ $+$ $1$

Substitute $x \,=\, 4$ on left hand side of the equation.

$=\,\,\,$ $\log{7}$ $+$ $\log{\big(3(4)-2\big)}$

$=\,\,\,$ $\log{7}$ $+$ $\log{(12-2)}$

$=\,\,\,$ $\log{7}$ $+$ $\log{10}$

$=\,\,\,$ $\log{7}$ $+$ $1$

Similarly, substitute the value of $x$ on right hand side of the equation.

$=\,\,\,$ $\log{(4+3)}$ $+$ $1$

$=\,\,\,$ $\log{7}$ $+$ $1$

Therefore, the values of both sides of the expressions are equal for $x$ equals to $4$. Hence, the value of $x$ equals to $4$ is solution for the given logarithmic equation.

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