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Answer 1

Q:Solve for X the equation 9^x - 3^x - 8 = 0

Solution
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=> 9^x - 3^x - 8 = 0
=> 9^x - 3^x = 8
=> 3 × 3^x - 3^x = 8
=> 3^x (3 - 1) = 8
=> 3^x × 2 = 8
=> 3^x = 8/2
=> 3^x = 4

Checking For 3^x = 4
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=> 3 × 4 - 4 - 8 = 0
=> 12 - 4 - 8 = 0
=> 8 - 8 = 0
=> 0 = 0

Hope It Helps!

Answer 2

Given Equation : $9^{x} - 3^{x} - 8 =0$


As 9 is the product of 3 and 3, it can be written as 3² .


$\therefore \: ( {3}^{2} ) {}^{x } - {3}^{x} - 8 = 0$

In the power of 3[ $(3^{2})^{x}$ ], 2 and x are in the product form, so it can be written as $(3^{x})^{2}$.

Now,


$\Rightarrow ( {3}^{x} ) {}^{2} - 3 {}^{x} - 8 = 0$


Let $3^{x} = a$


$\Rightarrow {a}^{2} - a - 8 = 0$


On Comparing the given equation with a²x + bx + c = 0 we get :

a = 1
b = - 1
c = - 8


$\Rightarrow a= \dfrac{ - ( - 1) \pm \sqrt{( - 1) {}^{2} - 4( - 8 \times 1)} }{2(1)} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg| \bold{by \: shridhara \: archaryyas \: method} \\ \\ \\ \Rightarrow a = \dfrac{1 \pm \sqrt{1 + 32} }{2} \\ \\ \\ \Rightarrow a = \frac{1 \pm \sqrt{33} }{2}$
$3 {}^{x} = \dfrac{1 \pm \: \sqrt{33} }{2}$


Neglecting negative value of $3^{x}$

taking log on both sides,


$\mathsf{ log_{3}( {3}^{x} ) = log_{3} \bigg( \dfrac{1 + \sqrt{33} }{2} \bigg) }$

From the properties of logarithms.


$x = \mathsf{ log_{3}\bigg( \dfrac{1 + \sqrt{33} }{2}\bigg)}$