Solve 9^n3^527^3/3*81=27

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

Solve for n

$$n=-3$$

Steps Using Definition of Logarithm

Cancel out 3 in both numerator and denominator.

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

Calculate 3 to the power of 4 and get 81.

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

Calculate 27 to the power of 3 and get 19683.

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

Multiply 81 and 19683 to get 1594323.

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

Divide 1594323\times 9^{n} by 81 to get 19683\times 9^{n}.

$$19683\times 9^{n}=27$$

Divide both sides by 19683.

$$9^{n}=\frac{27}{19683}$$

Reduce the fraction \frac{27}{19683} to lowest terms by extracting and canceling out 27.

$$9^{n}=\frac{1}{729}$$

Take the logarithm of both sides of the equation.

$$\log(9^{n})=\log(\frac{1}{729})$$

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

$$n\log(9)=\log(\frac{1}{729})$$

Divide both sides by \log(9).

$$n=\frac{\log(\frac{1}{729})}{\log(9)}$$

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

$$n=\log_{9}\left(\frac{1}{729}\right)$$