Solve 9^n*3^5*(27)^3/3*(81)^4=27

$$\frac { 9 ^ { n } \times 3 ^ { 5 } \times ( 27 ) ^ { 3 } } { 3 \times ( 81 ) ^ { 4 } } = 27$$

Solve for n

$$n=3$$

Steps Using Definition of Logarithm

Cancel out 3 in both numerator and denominator.

$$\frac{3^{4}\times 27^{3}\times 9^{n}}{81^{4}}=27$$

Calculate 3 to the power of 4 and get 81.

$$\frac{81\times 27^{3}\times 9^{n}}{81^{4}}=27$$

Calculate 27 to the power of 3 and get 19683.

$$\frac{81\times 19683\times 9^{n}}{81^{4}}=27$$

Multiply 81 and 19683 to get 1594323.

$$\frac{1594323\times 9^{n}}{81^{4}}=27$$

Calculate 81 to the power of 4 and get 43046721.

$$\frac{1594323\times 9^{n}}{43046721}=27$$

Divide 1594323\times 9^{n} by 43046721 to get \frac{1}{27}\times 9^{n}.

$$\frac{1}{27}\times 9^{n}=27$$

Multiply both sides by 27, the reciprocal of \frac{1}{27}.

$$9^{n}=27\times 27$$

Multiply 27 and 27 to get 729.

$$9^{n}=729$$

Take the logarithm of both sides of the equation.

$$\log(9^{n})=\log(729)$$

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$n\log(9)=\log(729)$$

Divide both sides by \log(9).

$$n=\frac{\log(729)}{\log(9)}$$

By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$n=\log_{9}\left(729\right)$$