Question

Solve the question. Find the roots of given equation. It's hard for me :(​

Answer 1

$\Huge\text{(a) -2}$

$\huge\text{\underline{\underline{Explanation}}}$

$\boxed{\text{Step 1.}}$

The first thing to do in these equations is to simplify. Via exponential laws, we can simplify.

$\cdots\longrightarrow 9^{x+2}-6\cdot3^{x+1}+1=0$

$\text{ \cdots\longrightarrow 9^{2}\cdot9^{x}-6\cdot3\cdot3^{x}+1=0\,\,\,\,\,\dots\boxed{a^{m}\times a^{n}=a^{m+n}} }$

$\cdots\longrightarrow 81\cdot9^{x}-18\cdot3^{x}+1=0$

So, we get-

$\cdots\longrightarrow 81\cdot3^{2x}-18\cdot3^{x}+1=0$

$\boxed{\text{Step 2.}}$

Now let's apply substitution to solve a quadratic equation. $(t=3^{x})$

$\cdots\longrightarrow 81t^2-18t+1=0$

$\cdots\longrightarrow (9t-1)^{2}=0$

$\cdots\longrightarrow t=\dfrac{1}{9}\text{ (Double solution)}$

$\boxed{\text{Step 3.}}$

Now, applying $t=3^{x}$-

$\cdots\longrightarrow 3^{x}=3^{-2}\text{ (Double solution)}$

$\cdots\longrightarrow x=-2\text{ (Double solution)}$

$\huge\text{\underline{\underline{Final answer}}}$

Hence, the equation has a double solution $x=-2$.

Option $\boxed{\text{(a) }-2}$ is correct.

For further questions ask in my questions.

Answer 2

Answer:

Option [A] -2

Step-by-step explanation:

Equation we are provided to solve,

$\longrightarrow \small\rm {9}^{x + 2} - 6 \cdot {3}^{x + 1} + 1 = 0$

Using laws of exponent, we get:

$\longrightarrow \small\rm {9}^{x} \cdot {9}^{2} - 6 \cdot {3}^{x} \cdot3 + 1 = 0$

$\longrightarrow \small\rm 81\cdot 3^{2x} - 18\cdot {3}^{x} + 1 = 0$

$\bf Let\, 3^x = y$

$\longrightarrow \small \rm 81\cdot y^{2} - 18\cdot y + 1 = 0$

This is quadratic equation in y. We can use quadratic formula to solve it.

By comparing this equation with general form of quadratic equation, we get:

• a = 81
• b = -18
• c = 1

Therefore, roots of equation are:

$\longmapsto\small \rm y = \dfrac{ - {b} \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

${\longmapsto\small \rm y = \dfrac{ - ( - 18) \pm \sqrt{ {( - 18)}^{2} - 4(81)(1)} }{2(81)}}$

${ \longmapsto\small \rm y = \dfrac{ 18\pm \sqrt{324 -324} }{162}}$

${\longmapsto\small \rm y = \dfrac{ 18 }{162}}$

${\longmapsto\small \rm y = \dfrac{ 1 }{9}}$

Substitute back the value of 7.

${\small \leadsto \rm {3}^{x} = \dfrac{ 1 }{9}}$

${\small \leadsto \rm {3}^{x} = \dfrac{ 1 }{ {3}^{2} }}$

${\small \leadsto \rm {3}^{x} = {3}^{ - 2} }$

$\pink{ \boxed{\small \leadsto \rm x = - 2 }}$

This is the required answer. Option [A] is correct.

Additional Information:-

$\boxed{\begin{array}{l }\:\underline{\text{Law of Exponents :}}\\\\\circ\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\ \circ\:\:\sf{(a^m)^n = a^{mn}}\\\\\circ\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\circ\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\circ\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{array}}$