Solve 1/243=81^5t-4

$$\frac { 1 } { 243 } = 81 ^ { 5 t - 4 }$$

Solve for t

$$t=\frac{11}{20}=0.55$$

Steps Using Definition of Logarithm

Swap sides so that all variable terms are on the left hand side.

$$81^{5t-4}=\frac{1}{243}$$

Take the logarithm of both sides of the equation.

$$\log(81^{5t-4})=\log(\frac{1}{243})$$

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

$$\left(5t-4\right)\log(81)=\log(\frac{1}{243})$$

Divide both sides by \log(81).

$$5t-4=\frac{\log(\frac{1}{243})}{\log(81)}$$

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

$$5t-4=\log_{81}\left(\frac{1}{243}\right)$$

Add 4 to both sides of the equation.

$$5t=-\frac{5}{4}-\left(-4\right)$$

Divide both sides by 5.

$$t=\frac{\frac{11}{4}}{5}$$