a type of plant is introduced into an ecosystem and quickly begins to take over. a scientist counts the…

a type of plant is introduced into an ecosystem and quickly begins to take over. a scientist counts the number of plants after m months and develops the equation (p(m)=19.3(1.089)^{m}) to model the situation. most recently, the scientist counted 138 plants. assuming there are no limiting factors to the growth of the plants, about how many months have passed since the plants were first introduced?\no 6.1\no 6.6\no 7.2\no 23.1

a type of plant is introduced into an ecosystem and quickly begins to take over. a scientist counts the number of plants after m months and develops the equation (p(m)=19.3(1.089)^{m}) to model the situation. most recently, the scientist counted 138 plants. assuming there are no limiting factors to the growth of the plants, about how many months have passed since the plants were first introduced?\no 6.1\no 6.6\no 7.2\no 23.1

Answer

Explanation:

Step1: Set up the equation

We know that $P(m)=138$ and $P(m) = 19.3(1.089)^m$, so $138=19.3(1.089)^m$.

Step2: Isolate the exponential term

Divide both sides by 19.3: $\frac{138}{19.3}=(1.089)^m$, which simplifies to approximately $7.15026=(1.089)^m$.

Step3: Take the natural - logarithm of both sides

$\ln(7.15026)=\ln(1.089^m)$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(7.15026)=m\ln(1.089)$.

Step4: Solve for m

$m = \frac{\ln(7.15026)}{\ln(1.089)}$. Since $\ln(7.15026)\approx1.967$ and $\ln(1.089)\approx0.0852$, then $m=\frac{1.967}{0.0852}\approx23.1$.

Answer:

23.1