Question

If ${{3}^{x}}+27\left( {{3}^{-x}} \right)=12$, then what is the value of $x$?
(a) 1 only
(b) 2 only
(c) 1 or 2
(d) 0 or 1

Answer 1

Hint: We must first assume a variable $y={{3}^{x}}$, and then solve to get a quadratic equation in y. We can find the two roots of y and using them, we can find the values of variable x, which will be our solution.

Complete step-by-step answer:
We all know very well that ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$. Thus, by using this property, we can write ${{3}^{-x}}=\dfrac{1}{{{3}^{x}}}$.
Hence, our equation now becomes ${{3}^{x}}+\dfrac{27}{{{3}^{x}}}=12$.
Let us now assume a variable $y={{3}^{x}}$.
Thus, we have $y+\dfrac{27}{y}=12$.
Let us simplify the right hand side of this equation by taking LCM. Thus, we get
$\dfrac{{{y}^{2}}+27}{y}=12$
So, we can also write
${{y}^{2}}+27=12y$.
Or, in standard form
${{y}^{2}}-12y+27=0$.
We know that any equation of the form $a{{y}^{2}}+by+c=0$, is called a quadratic equation in y and will have two solutions.
Let us break the middle term 12y into two parts 9y and 3y. Thus, we get
${{y}^{2}}-9y-3y+27=0$.
Taking y as common from the first two terms and 3 as common from the last two terms on the left hand side, we get
$y\left( y-9 \right)-3\left( y-9 \right)=0$.
Hence, we can write the above equation as,
$\left( y-9 \right)\left( y-3 \right)=0$
Now, either (y – 9) = 0 or (y – 3) = 0. Thus, we will have two cases.
Case (i): $\left( y-9 \right)=0$, thus we get $y=9$.
Hence, ${{3}^{x}}=9$, or we can say that $x=2$.
Case (ii): $\left( y-3 \right)=0$, thus we get $y=3$.
Hence, ${{3}^{x}}=3$, or we can say that $x=1$.
Thus, the value of $x$ is 1 or 2.

So, the correct answer is “Option (c)”.

Note: We can also solve this problem by following an objective approach, in which we can use ${{3}^{3}}=27$, and substituting the options one by one into the resulting equation. We must note that we can use this approach only for questions that doesn’t have ‘None of these’ as an option.