Respuesta :

The trick is to express everything in terms of powers of 3 (since both 9 and 81 are powers of 3):

[tex]9=3^2,\quad 81=3^4[/tex]

So, the equation becomes

[tex](3^2)^{2x+1}=\dfrac{(3^4)^{x-2}}{3^x}[/tex]

Apply the power rule [tex](a^b)^c=a^{bc}[/tex] to get

[tex]3^{4x+2}=\dfrac{3^{4x-8}}{3^x}[/tex]

And finally the rule [tex]\frac{a^b}{a^c}=a^{b-c}[/tex] to get

[tex]3^{4x+2}=3^{3x-8}[/tex]

Now we get to the simple part: two powers of the same base are equal if and only if the exponents equal each other:

[tex]4x+2=3x-8 \iff x=-10[/tex]

lavanyaande

The value of x is -10.

Step-by-step explanation:

Given,

[tex]9^{2x+1} = \frac{81^{x-2} }{3^{x} }[/tex]

To find value of x.

Formula

[tex]\frac{a^{m} }{a^{n} } =a^{m-n}[/tex]

Now,

[tex]9^{2x+1} = \frac{81^{x-2} }{3^{x} }[/tex]

or, [tex]3^{2(2x+1)} = \frac{3^{4(x-2)} }{3^{x} }[/tex]

or, [tex]3^{2(2x+1)} = 3^{4(x-2)-x}[/tex]

or, 2(2x+1) = 4(x-2)-x [ since the base is equal the powers are also equal]

or, 4x+2 = 4x-8-x

or, 4x-4x+x = -8-2

or, x = -10

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