Question

Let ${{3}^{a}}=4,{{4}^{b}}=5,{{5}^{c}}=6,{{6}^{d}}=7,{{7}^{e}}=8$ and ${{8}^{f}}=9$. The value of the product $\left( abcdef \right)$ is
$\begin{align}
  & \text{A}\text{. 2} \\
 & \text{B}\text{. 1} \\
 & \text{C}\text{. }\sqrt{6} \\
 & \text{D}\text{. 3} \\
\end{align}$

Answer 1

Hint:To solve this question we use the property ${{\left( {{a}^{n}} \right)}^{m}}={{a}^{mn}}$. We substitute the values from the given equations and at last compare the exponents of the obtained equation to find the product of $\left( abcdef \right)$.

Complete step by step answer:
We have been given ${{3}^{a}}=4,{{4}^{b}}=5,{{5}^{c}}=6,{{6}^{d}}=7,{{7}^{e}}=8$ and ${{8}^{f}}=9$.
We have to find the value of the product $\left( abcdef \right)$.
Now, let us consider ${{3}^{a}}=4..........(i)$and ${{4}^{b}}=5..........(ii)$
Now, we substitute the value of $4$ from equation (i) into equation (ii), we get
${{\left( {{3}^{a}} \right)}^{b}}=5$
Now, we know that ${{\left( {{a}^{n}} \right)}^{m}}={{a}^{mn}}$
So, we have ${{3}^{ab}}=5...........(iii)$
Now, we have ${{5}^{c}}=6$
Now, we substitute the value of $5$ from equation (iii), we get
${{\left( {{3}^{ab}} \right)}^{c}}=6$
$\Rightarrow \left( {{3}^{abc}} \right)=6............(iv)$
Now, we have ${{6}^{d}}=7$
Now, we substitute the value of $6$ from equation (iv), we get
${{\left( {{3}^{abc}} \right)}^{d}}=7$
Or we can write that $\left( {{3}^{abcd}} \right)=7...............(v)$
Now, we have ${{7}^{e}}=8$
Now, we substitute the value of $7$ from equation (v), we get
${{\left( {{3}^{abcd}} \right)}^{e}}=8$
Or $\left( {{3}^{abcde}} \right)=8.............(vi)$
Now, we have ${{8}^{f}}=9$
Now, we substitute the value of $8$ from equation (vi), we get
${{\left( {{3}^{abcde}} \right)}^{f}}=9$
Or ${{3}^{abcdef}}=9$
Now, we know that $9={{3}^{2}}$
So, we have ${{3}^{abcdef}}={{3}^{2}}$
Now, we have same base from both sides of equation, if we compare exponents we get
$abcdef=2$
So the value of the product $\left( abcdef \right)$ is $2$.
So, the correct answer is “Option A”.

Note:
 To solve such type of questions students must have knowledge of exponentiation rules. The possibility of mistake is while substituting the values and solving the exponents. Be careful while substituting values, one incorrect step leads to incorrect solution.
We have ${{3}^{a}}=4$ and ${{4}^{b}}=5$, if students try to substitute the value of $4$ from second equation then, it gives incorrect solution.