log5 v+log(1/3) 9+log9 81 = - 4. To solve for v.

log (5) v +log(1/3) 9+log(9) 81 = -4.

log(5) v + log(1/3) (1/3) ^(-2) + log(9) 9^2 = -4.

log(5) v - 2 + 2 = -4, as log (a) a^m = m. So log (1/3) 9 = log(1/3) (1/3)^-2 = -2 and log (9) 81 = log (9) 9^2 = 2.

log(5) v = -4.

v = 5^(-4) = 1/5^4 = 1/625.

Therefore v = 1/625.

We'll identify v as the unknown that has to be found.

We'll isolate the unknown to the left side and we'll move the rest of the terms to the right side:

log5 v = -4 - log9 81 - log(1/3) 9

We'll apply the power rule of logarithms for the term:

log9 81 = log9 (9^2) = 2log9 9 = 2

We'll change the base of the term log(1/3) 9

log(1/3) 9 = 1/log9 (1/3)

log3 (1/3) = log9 (1/3)*log3 9

log9 (1/3) = log3 (1/3)/log3 9

We'll re-write the numerator and denominator:

log3 (1/3)= log3 (3^-1) = -1

log3 9 = log3 (3^2) = 2

log9 (1/3) = -1/2

log5 v = -4 - 2 - 1/(-1/2)

log5 v = -4 - 2 + 2

We'll eliminate like terms:

log5 v = -4

W'll take antilogarithms and we'll get:

v = 5^-4

v = 1/5^4

**v = 1/625**