Question

If ${3^x} = z$, ${z^y} = 5$, ${5^z} = 3$ then $xyz =$?
(A) 0
(B) 5
(C) 3
(D) 1

Answer 1

The law of exponents states that when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. For any real numbers $a$, $x$ and $y$, where $a > 0$, ${x} = {y}$ if and only if ${a^x} = {a^y}$.Using this concept we try to solve the question.

Complete step-by-step answer:
Given equations are,
${3^x} = z ….…. (1)$
${z^y} = 5 ….…. .(2)$
${5^z} = 3 ...…. (3)$
Substitute the value of $z$ from equation (1) to equation (2),
${\left( {{3^x}} \right)^y} = 5$
$\Rightarrow {3^{xy}} = 5 …..…. (4)$
Now, substitute the value of $5$ from equation (4) to equation (3),
${\left( {{3^{xy}}} \right)^z} = 3$
$\Rightarrow {3^{xyz}} = {3^1}$ …. (5)
As we know that when an exponential equation has the same base on each side, the exponents must be equal.
Since in equation (5), the base on both sides is the same i.e., 3. Therefore, the exponents also are equal.
$\Rightarrow xyz = 1$

So, the correct answer is “Option D”.

Note: We can also solve this question by an alternative method.
In this method, we multiplied all the given equations (1), (2) and (3);
${3^x} \cdot {z^y} \cdot {5^z} = z \times 5 \times 3$
$\Rightarrow$ $\dfrac{{{3^x} \cdot {z^y} \cdot {5^z}}}{{z \times 5 \times 3}} = 1$
$\Rightarrow$ ${3^{x - 1}} \cdot {z^{y - 1}} \cdot {5^{z - 1}} = 1$
As we know that ${a^0} = 1$. By using this concept, the above equation can be written as-
$\Rightarrow$ ${3^{x - 1}} \cdot {z^{y - 1}} \cdot {5^{z - 1}} = {3^0} \cdot {z^0} \cdot {5^0}$
The law of exponents states that when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. For any real numbers $a$, $x$ and $y$, where $a > 0$, ${x} = {y}$ if and only if ${a^x} = {a^y}$.
So comparing the power of both sides with same base,
$x - 1 = 0 \Rightarrow x = 1$
$y - 1 = 0 \Rightarrow y = 1$
$z - 1 = 0 \Rightarrow z = 1$
Now, $xyz = 1 \times 1 \times 1 = 1$