$a$, $b$ and $c$ are three literals. They have a mathematical relation through common logarithm and ratio. The relation between them are given here.

$\dfrac{\log a}{b-c}$ $\,=\,$ $\dfrac{\log b}{c-a}$ $\,=\,$ $\dfrac{\log c}{a-b}$

The value of the product $a$, $b$ and $c$ is required to find by using this data.

01

$abc$ is a product of three literals $a$, $b$ and $c$. Fortunately, there is a logarithmic identity to express the product of the terms as sum of their logs. So, use the product rule of logarithms and express log of product of $a$, $b$ and $c$ as sum of their logs.

$\log abc$ $\,=\,$ $\log a + \log b + \log c$

02

According to the given data of this problem. every logarithmic term can be transformed into another logarithmic term. So, let us transform the first two terms in terms of the $\log c$.

$\dfrac{\log a}{b-c} = \dfrac{\log c}{a-b}$

$\implies \log a = (b-c) \times \dfrac{\log c}{a-b}$

$\therefore \,\,\,\,\,\, \log a = \dfrac{b-c}{a-b} \times \log c$

In the same way,

$\dfrac{\log b}{c-a} = \dfrac{\log c}{a-b}$

$\implies \log b = (c-a) \times \dfrac{\log c}{a-b}$

$\therefore \,\,\,\,\,\, \log b = \dfrac{c-a}{a-b} \times \log c$

Now, replace the terms $\log a$ and $\log b$ in the logarithmic equation.

$\implies$ $\log abc$ $\,=\,$ $\dfrac{b-c}{a-b} \times \log c$ $+$ $\dfrac{c-a}{a-b} \times \log c$ $+$ $\log c$

$\log c$ is a common factor in the three terms of the equation. So, take it common from them.

$\implies$ $\log abc$ $\,=\,$ $\log c \times \Bigg[\dfrac{b-c}{a-b} + \dfrac{c-a}{a-b} + 1\Bigg]$

$\implies$ $\log abc$ $\,=\,$ $\log c \times \Bigg[\dfrac{b-c + c-a + a-b}{a-b}\Bigg]$

$\implies$ $\log abc$ $\,=\,$ $\require{cancel} \log c \times \Bigg[\dfrac{\cancel{b} -\cancel{c} + \cancel{c} -\cancel{a} +\cancel{a} -\cancel{b}}{a-b}\Bigg]$

$\implies$ $\log abc$ $\,=\,$ $\log c \times \Bigg[\dfrac{0}{a-b}\Bigg]$

$\implies$ $\log abc \,=\, \log c \times 0$

$\implies$ $\log abc \,=\, 0$

03

In order to obtain the value of $abc$, write the equation $\log abc = 0$ in exponential form as per the mathematical relation between logarithms and exponents.

$\implies$ $abc \,=\, 10^0$

Use power rule of zero exponent to evaluate $10$ raised to the power of zero.

$\therefore \,\,\,\,\,\,$ $abc \,=\, 1$

Therefore, the product of the literals $a$, $b$ and $c$ is $1$ and it is the required solution for this logarithmic problem.

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