Takumi plants a tree in his backyard and studies how the number of branches grows over time. $\newline$ He predicts that the relationship between $N$ , the number of branches on the tree, and $t$ , the elapsed time, in years, since the tree was planted can be modeled by the following equation. $\newline$ $N=5 \cdot 10^{0.3 t}$ $\newline$ According to Takumi's model, in how many years will the tree have $100$ branches? $\newline$ Give an exact answer expressed as a base $-10$ logarithm. $\newline$ years

Full solution

Q. Takumi plants a tree in his backyard and studies how the number of branches grows over time.

\newline

He predicts that the relationship between

N

, the number of branches on the tree, and

t

, the elapsed time, in years, since the tree was planted can be modeled by the following equation.

\newline

N=5 \cdot 10^{0.3 t}

\newline

According to Takumi's model, in how many years will the tree have

100

branches?

\newline

Give an exact answer expressed as a base

-10

logarithm.

\newline

years

Write Equation and Value: Write down the given equation and the value for $N$ that we want to achieve. $\newline$ We are given the equation $N = 5 \times 10^{(0.3t)}$ and we want to find the value of $t$ when $N = 100$ .
Set Up and Solve: Set up the equation with $N = 100$ and solve for $t$ . $100 = 5 \times 10^{(0.3t)}$
Isolate Exponential Term: Divide both sides of the equation by $5$ to isolate the exponential term. $\newline$ $100 / 5 = 10^{0.3t}$ $\newline$ $20 = 10^{0.3t}$
Apply Logarithm: Apply the logarithm to both sides of the equation to solve for $t$ . $\log(20) = \log(10^{0.3t})$
Simplify Equation: Use the property of logarithms that $\log(a^b) = b \cdot \log(a)$ to simplify the right side of the equation. $\log(20) = 0.3t \cdot \log(10)$
Divide and Solve: Since $\log(10)$ is $1$ , we can simplify the equation further. $\newline$ $\log(20) = 0.3t$
Final Answer: Divide both sides of the equation by $0.3$ to solve for $t$ . $\newline$ $t = \frac{\log(20)}{0.3}$
Final Answer: Divide both sides of the equation by $0.3$ to solve for $t$ . $\newline$ $t = \frac{\log(20)}{0.3}$ Express the final answer as a base- $10$ logarithm. $\newline$ $t = \frac{\log(20)}{\log(10^{0.3})}$