Question

This equation is hard to solve. please leave your working out.

find x​

Answer 1

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

`9=3^2,\quad 81=3^4`

So, the equation becomes

`(3^2)^(2x+1)=((3^4)^(x-2))/(3^x)`

Apply the power rule
`(a^b)^c=a^(bc)` to get

`3^(4x+2)=(3^(4x-8))/(3^x)`

And finally the rule
`(a^b)/(a^c)=a^(b-c)` to get

`3^(4x+2)=3^(3x-8)`

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

`4x+2=3x-8 \iff x=-10`

Answer 2

The value of x is -10.

Explanation:

Given,

`9^(2x+1) = (81^(x-2) )/(3^(x) )`

To find value of x.

Formula

`(a^(m) )/(a^(n) ) =a^(m-n)`

Now,

`9^(2x+1) = (81^(x-2) )/(3^(x) )`

or,
`3^(2(2x+1)) = (3^(4(x-2)) )/(3^(x) )`

or,
`3^(2(2x+1)) = 3^(4(x-2)-x)`

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