Question

Solve for ‘x’, ${{9}^{x+2}}-{{6.3}^{x+2}}+1=0$?

Answer 1

Hint: We start solving the problem by applying the law of exponents ${{a}^{m+n}}={{a}^{m}}.{{a}^{n}}$ in the given equation. We then apply the law of exponents ${{\left( {{a}^{m}} \right)}^{n}}={{\left( {{a}^{n}} \right)}^{m}}$ and assume the ${{3}^{x}}=y$ to get a quadratic equation in ‘y’. We then find the roots of the obtained quadratic equation in ‘y’ and equate it to ${{3}^{x}}$. We then make use of laws of exponents $\dfrac{1}{{{a}^{x}}}={{a}^{-x}}$ and if ${{a}^{m}}={{a}^{n}}$ then $m=n$ to get the required value of ‘x’.

Complete step by step answer:
According to the problem we need to find the value of x which satisfies the equation ${{9}^{x+2}}-{{6.3}^{x+1}}+1=0$.
So, we have ${{9}^{x+2}}-{{6.3}^{x+1}}+1=0$.
From the laws of exponents, we know that ${{a}^{m+n}}={{a}^{m}}.{{a}^{n}}$.
So, we get ${{9}^{2}}{{.9}^{x}}-{{6.3}^{1}}{{.3}^{x}}+1=0$.
$\Rightarrow {{81.9}^{x}}-{{6.3.3}^{x}}+1=0$.
$\Rightarrow 81.{{\left( {{3}^{2}} \right)}^{x}}-{{18.3}^{x}}+1=0$.
From the law of exponents, we know that ${{\left( {{a}^{m}} \right)}^{n}}={{\left( {{a}^{n}} \right)}^{m}}$.
$\Rightarrow 81.{{\left( {{3}^{x}} \right)}^{2}}-{{18.3}^{x}}+1=0$.
Let us assume ${{3}^{x}}=y$ ---(1). So, we get $81{{y}^{2}}-18y+1=0$.
We can see that $81{{y}^{2}}-18y+1=0$ resembles a quadratic equation $a{{x}^{2}}+bx+c=0$. Let us factorize it and find the roots.
$\Rightarrow 81{{y}^{2}}-9y-9y+1=0$.
$\Rightarrow 9y\left( 9y-1 \right)-1\left( 9y-1 \right)=0$.
$\Rightarrow \left( 9y-1 \right)\left( 9y-1 \right)=0$.
$\Rightarrow {{\left( 9y-1 \right)}^{2}}=0$.
$\Rightarrow 9y-1=0$.
$\Rightarrow 9y=1$.
$\Rightarrow y=\dfrac{1}{9}$. Let us substitute this in equation (1).
$\Rightarrow {{3}^{x}}=\dfrac{1}{9}$.
$\Rightarrow {{3}^{x}}=\dfrac{1}{{{3}^{2}}}$.
From the law of exponents, we know that $\dfrac{1}{{{a}^{x}}}={{a}^{-x}}$.
$\Rightarrow {{3}^{x}}={{3}^{-2}}$.
From the law of exponents, we know that if ${{a}^{m}}={{a}^{n}}$ then $m=n$.
$\therefore x=-2$.

So, we have found the value of ‘x’ as –2.

Note: Whenever we get this type of problems, we try to assume a variable for the term with the independent variable in exponent to avoid confusion while solving this problem. We should not make mistakes while applying the laws of exponents in this problem. We can also solve for the roots of the quadratic equation $81{{y}^{2}}-18y+1=0$ by applying $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Similarly, we can expect problems to find the value of ${{5}^{x+2}}$ after finding the value of ‘x’.