Math Doubts

Solve $\log 7+\log(3x-2)$ $=$ $\log(x+3)+1$ and find the value of $x$

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

There is no base for logarithmic term in the given math question. So, it is known as a common logarithmic system. Therefore, its base is $10$.

In the given logarithmic problem, the logarithmic terms are added each other in left hand side of the equation. They can be converted as a single logarithmic term by using the product rule of logarithms.

One term is logarithmic term and other is number in right hand side of the equation. If the number is transformed as a logarithmic term, the right hand side equation also consists of two logarithmic terms and they can be combined as a single logarithmic term. According to fundamental laws of logarithms, the logarithm of $10$ to base $10$ is $1$. So, the number $1$ can be written as $\log 10$.

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

Now, use product law of logarithm and combine both left hand side and right hand sides of the equations as single logarithmic terms.

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

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

This mathematical statement clears that logarithm of an expression is equal to logarithmic of another expression. Therefore, they two algebraic equations should be equal mathematically.

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

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

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

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

$\implies 11x = 44$

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

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

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

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


Substitute $x = 4$ in left hand side of the equation.

$\log 7+\log(3x-2) = \log 7+\log(3(4)-2)$

$\implies \log 7+\log(3x-2) = \log 7+\log(12-2)$

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

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

$\therefore \,\,\,\,\, \log 7+\log(3x-2) = \log 70$

Now, substitute the value of $x$ in right hand side of the equation.

$\log(x+3)+1 = \log(4+3)+1$

$\implies \log(x+3)+1 = \log 7 + \log 10$

$\implies \log(x+3)+1 = \log (7 \times 10)$

$\therefore \,\,\,\,\, \log(x+3)+1 = \log 70$

It is proved that, for $x=4$

$\log 7+\log(3x-2) = \log(x+3)+1 = \log 70$

Therefore, It is confirmed that the value of $x$ is $4$.

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