Find the value of $\log_{d} abc$ if $a^2 \,=\, b^3 \,=\, c^5 \,=\, d^6$

$a$, $b$, $c$ and $d$ are four literals. The square of $a$, cube of $b$, fifth power of $c$ and sixth power of $d$ are equal.

$a^2 \,=\, b^3 \,=\, c^5 \,=\, d^6$, then the value of logarithm of product of $a$, $b$ and $c$ to base $d$ is required to find and it is written in logarithms as $\displaystyle \log_{d} abc$ mathematically.

The logarithm of product of literals $a$, $b$ and $c$ is evaluated on the basis of literal number $d$. Hence, transform the exponents of literal numbers $a$, $b$ and $c$ in terms of $d$ purely.

$a^2 \,=\, d^6 \,\implies\, a \,=\, {(d^6)}^{\dfrac{1}{2}}$

$b^3 \,=\, d^6 \,\implies\, b \,=\, {(d^6)}^{\dfrac{1}{3}}$

$c^5 \,=\, d^6 \,\implies\, c \,=\, {(d^6)}^{\dfrac{1}{5}}$

Substitute the value of $a$, $b$ and $c$ in terms $d$ in the logarithmic term.

$\displaystyle \log_{d} abc$ $\,=\,$ $\displaystyle \log_{d} \Big[{(d^6)}^{\dfrac{1}{2}} {(d^6)}^{\dfrac{1}{3}} {(d^6)}^{\dfrac{1}{4}}\Big]$

$d^6$ is an exponential term and it is a common base for the three exponential terms. So, use the product rule of exponents to simplify it.

$=\,\,\,$ $\log_{d} {(d^6)}^{\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{5}}$

$=\,\,\,$ $\log_{d} {(d^6)}^{\dfrac{15+10+6}{30}}$

$=\,\,\,$ $\log_{d} {(d^6)}^{\dfrac{31}{30}}$

The exponential term also has another exponent. It can be simplified by the power rule of power of an exponential term.

$=\,\,\,$ $\log_{d} d^{\displaystyle 6 \times \dfrac{31}{30}}$

$=\,\,\,$ $\log_{d} d^{\dfrac{6 \times 31}{30}}$

$=\,\,\,$ $\require{cancel} \log_{d} d^{\dfrac{\cancel{6} \times 31}{\cancel{30}}}$

$=\,\,\,$ $\log_{d} d^{\dfrac{31}{5}}$

Use power rule of logarithms to simplify it further.

$=\,\,\,$ $\dfrac{31}{5} \times \log_{d} d$

According to logarithm of base rule, the value is one.

$=\,\,\,$ $\dfrac{31}{5} \times 1$

$=\,\,\,$ $\dfrac{31}{5}$