Question

If $x = {\log _3}15,$$y = {\log _5}3,$$z = {\log _{15}}5,$ then find the value of $xyz.$

Answer 1

Hint: Simplify the question by making their bases equal. It would help to get a proper approach for the question.

Step by step solution:
It is given that
$x = {\log _3}15$
$y = {\log _5}3$
$z = {\log _{15}}5$
First, we will make sure that all the bases are equal.
We know,
${\log _a}b = \dfrac{1}{{{{\log }_b}a}}$
Using this, we can write
$y = {\log _5}3 = \dfrac{1}{{{{\log }_3}5}}$ . . . . (1)
And $z = {\log _{15}}5 = \dfrac{1}{{{{\log }_5}15}}$
We know , ${\log _a}(m \times n) = {\log _a}m + {\log _a}n$
Using this, we can write
$x = {\log _3}(5 \times 3)$
$ = {\log _3}5 + {\log _3}3$
$ \Rightarrow x = {\log _3}5 + 1$ . . . . . (2)
$z = \dfrac{1}{{{{\log }_5}15}} = \dfrac{1}{{{{\log }_5}(3 \times 5)}}$
$ = \dfrac{1}{{{{\log }_5}3 + {{\log }_5}5}}$
$z = \dfrac{1}{{\dfrac{1}{{{{\log }_3}5}} + 1}}$ . . . . (3) $\left( {\because {{\log }_a}a = 1,{{\log }_a}b = \dfrac{1}{{{{\log }_b}a}}} \right)$
Now, we have to find the value of $xyz$. For that we put the values of $y,x$ and $z$ from equations (1), (2) and (3) respectively.
$xyz = ({\log _3}5 + 1)\left( {\dfrac{1}{{{{\log }_3}5}}} \right)\left( {\dfrac{1}{{\dfrac{1}{{{{\log }_3}5}} + 1}}} \right)$
To make it simple let us put ${\log _3}5 = t$
$xyz = (t + 1) \times \dfrac{1}{t} \times \dfrac{1}{{\dfrac{1}{t} + 1}}$
$ = \dfrac{{t + 1}}{t} \times \dfrac{1}{{\dfrac{{(1 + t)}}{t}}}$
$ = \dfrac{{t + 1}}{t} \times \dfrac{t}{{1 + t}}$
$xyz = 1$
Therefore the value of $xyz = 1$

Note: Always reduce logarithmic terms to simple forms until we get equal base. Substitute the repeating term to simplify the question.