Given:

`(\frac{1}{9})^m=81^{m+4}`

We have to find the value of m.

So we can write,

`9^{-m}=(9^2)^{m+4}`

i.e. `9^{-m}=9^{2m+8}`

Comparing the left hand side and the right hand side we get,

`-m=2m+8`

i.e. `-3m=8`

which means `m=\frac{-8}{3}`

To solve this, first make both bases the same. 1/9^m is equivalent to 9^-m, and 9^2 equals 81 so the equation becomes:

`9^-m=9^(2(m+4))`

Cancel out the bases since they're the same number.

-m=2(m+4)

-m=2m+8

-8=3m

m=-8/3 This is the answer. You can also convert this into a decimal.

`(1/9)^m = 81^(m+4)`

The first step to make this easier is to simplify the problems so that they have the same base.

We know that in order to make 1/9 a 9 we have to have a - for m because then we use the reciprocal

`(1/9)^m =9^-m`

81 can be simplified as 9^2 therefore

`9^-m = 9^(2(m+4))`

Now set the exponent equal and solve:

Distribute the 2

`-m = 2m + 8`

move like terms to the same side:

`-3m = 8`

Divide by -3 to get m alone

`m = -8/3`

`-8/3` is the answer

to check

`(1/9)^(-8/3) =350.5 `

`81^((-8/3 + 4)) = 350.5`