Question

For which value of $a,3^{-3a}=\frac{1}{27}?$

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $1$
• D. None of the above

Answer 1

The correct option is C $1$

$\underset{–}{\text{Need to find:}}$ Value of $a$ which satisfies the exponential equation $3^{-3a}=\frac{1}{27}$


$3^{-3a}=\frac{1}{27}$

Applying $a^{-n}=\frac{1}{a^n}:$

$\Rightarrow \frac{1}{3^{3a}}=\frac{1}{3^3}$ $[\because 27=3\times 3\times 3=3^3]$

$\Rightarrow \frac{1}{3^{3a}}\times 3^3=\frac{1}{3^3}\times 3^3$
[Multiplying $3^3$ both sides]

$\Rightarrow \frac{3^3}{3^{3a}}=1$

​​​​​​​$\Rightarrow 3^{3-3a}=3^0$
$(\because \frac{a^m}{a^n}=a^{m-n}\text{and}{p}^0=1\text{for all}p\ne 0)$

In the above exponential equation, the base $\left(3\right)$ is same. Hence, for L.H.S=R.H.S, the powers must be same.

$\Rightarrow 3-3a=0$
$\Rightarrow 3-3a+3a=0+3a$
$\Rightarrow 3=3a$
$\Rightarrow 3a=3$
$\Rightarrow \frac{3a}{3}=\frac{3}{3}$
$\Rightarrow \underset{–}{a=1}$

$\therefore$ The value of $a$ which satisfies the exponential equation $3^{-3a}=\frac{1}{27}$ is $\underset{–}{1}$.

Hence, option (c.) is the correct choice.