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If $\dfrac{\log a}{b-c}$ $\,=\,$ $\dfrac{\log b}{c-a}$ $\,=\,$ $\dfrac{\log c}{a-b}$, find the value of $abc$

$a$, $b$ 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

Application of Product Rule

$abc$ 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

Transform any two terms into remaining term

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

Transformation of the logarithmic term

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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