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Question: Solve for ‘x’, \({{9}^{x+2}}-{{6.3}^{x+2}}+1=0\)?...

Solve for ‘x’, 9x+26.3x+2+1=0{{9}^{x+2}}-{{6.3}^{x+2}}+1=0?

Explanation

Solution

We start solving the problem by applying the law of exponents am+n=am.an{{a}^{m+n}}={{a}^{m}}.{{a}^{n}} in the given equation. We then apply the law of exponents (am)n=(an)m{{\left( {{a}^{m}} \right)}^{n}}={{\left( {{a}^{n}} \right)}^{m}} and assume the 3x=y{{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 3x{{3}^{x}}. We then make use of laws of exponents 1ax=ax\dfrac{1}{{{a}^{x}}}={{a}^{-x}} and if am=an{{a}^{m}}={{a}^{n}} then m=nm=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 9x+26.3x+1+1=0{{9}^{x+2}}-{{6.3}^{x+1}}+1=0.
So, we have 9x+26.3x+1+1=0{{9}^{x+2}}-{{6.3}^{x+1}}+1=0.
From the laws of exponents, we know that am+n=am.an{{a}^{m+n}}={{a}^{m}}.{{a}^{n}}.
So, we get 92.9x6.31.3x+1=0{{9}^{2}}{{.9}^{x}}-{{6.3}^{1}}{{.3}^{x}}+1=0.
81.9x6.3.3x+1=0\Rightarrow {{81.9}^{x}}-{{6.3.3}^{x}}+1=0.
81.(32)x18.3x+1=0\Rightarrow 81.{{\left( {{3}^{2}} \right)}^{x}}-{{18.3}^{x}}+1=0.
From the law of exponents, we know that (am)n=(an)m{{\left( {{a}^{m}} \right)}^{n}}={{\left( {{a}^{n}} \right)}^{m}}.
81.(3x)218.3x+1=0\Rightarrow 81.{{\left( {{3}^{x}} \right)}^{2}}-{{18.3}^{x}}+1=0.
Let us assume 3x=y{{3}^{x}}=y ---(1). So, we get 81y218y+1=081{{y}^{2}}-18y+1=0.
We can see that 81y218y+1=081{{y}^{2}}-18y+1=0 resembles a quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0. Let us factorize it and find the roots.
81y29y9y+1=0\Rightarrow 81{{y}^{2}}-9y-9y+1=0.
9y(9y1)1(9y1)=0\Rightarrow 9y\left( 9y-1 \right)-1\left( 9y-1 \right)=0.
(9y1)(9y1)=0\Rightarrow \left( 9y-1 \right)\left( 9y-1 \right)=0.
(9y1)2=0\Rightarrow {{\left( 9y-1 \right)}^{2}}=0.
9y1=0\Rightarrow 9y-1=0.
9y=1\Rightarrow 9y=1.
y=19\Rightarrow y=\dfrac{1}{9}. Let us substitute this in equation (1).
3x=19\Rightarrow {{3}^{x}}=\dfrac{1}{9}.
3x=132\Rightarrow {{3}^{x}}=\dfrac{1}{{{3}^{2}}}.
From the law of exponents, we know that 1ax=ax\dfrac{1}{{{a}^{x}}}={{a}^{-x}}.
3x=32\Rightarrow {{3}^{x}}={{3}^{-2}}.
From the law of exponents, we know that if am=an{{a}^{m}}={{a}^{n}} then m=nm=n.
x=2\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 81y218y+1=081{{y}^{2}}-18y+1=0 by applying b±b24ac2a\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}. Similarly, we can expect problems to find the value of 5x+2{{5}^{x+2}} after finding the value of ‘x’.