Evaluate 5*2^(-3t)=45

$5 \cdot 2^{- 3 t} = 45$

Step 1

Divide each term in $(5 \cdot 2^{- 3 t}) = 45$ by $5$ and simplify.

Divide each term in $(5 \cdot 2^{- 3 t}) = 45$ by $5$ .

$\frac{(5 \cdot 2^{- 3 t})}{5} = \frac{45}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 \cdot 2^{- 3 t}}{5} = \frac{45}{5}$

Divide $2^{- 3 t}$ by $1$ .

$2^{- 3 t} = \frac{45}{5}$

Simplify the right side.

Divide $45$ by $5$ .

$2^{- 3 t} = 9$

Step 2

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (2^{- 3 t}) = \ln (9)$

Step 3

Expand $\ln (2^{- 3 t})$ by moving $- 3 t$ outside the logarithm.

$- 3 t \ln (2) = \ln (9)$

Step 4

Divide each term in $- 3 t \ln (2) = \ln (9)$ by $- 3 \ln (2)$ and simplify.

Divide each term in $- 3 t \ln (2) = \ln (9)$ by $- 3 \ln (2)$ .

$\frac{- 3 t \ln (2)}{- 3 \ln (2)} = \frac{\ln (9)}{- 3 \ln (2)}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 t \ln (2)}{- 3 \ln (2)} = \frac{\ln (9)}{- 3 \ln (2)}$

Rewrite the expression.

$\frac{t \ln (2)}{\ln (2)} = \frac{\ln (9)}{- 3 \ln (2)}$

Cancel the common factor of $\ln (2)$ .

Cancel the common factor.

$\frac{t \ln (2)}{\ln (2)} = \frac{\ln (9)}{- 3 \ln (2)}$

Divide $t$ by $1$ .

$t = \frac{\ln (9)}{- 3 \ln (2)}$

Simplify the right side.

Move the negative in front of the fraction.

$t = - \frac{\ln (9)}{3 \ln (2)}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$t = - \frac{\ln (9)}{3 \ln (2)}$

Decimal Form:

$t = - 1.05664166 \dots$