Yasemin deposited $1000 into a savings account. The relationship between the time, t, in years, since the account was first opened, and Yasemin's account balance, B(t), in dollars, is modeled by the following function. B(t)=1000*e^(0.03 t) How many years will it take for Yasemin's account balance to reach $1500 ? Round your answer, if necessary, to the nearest hundredth. years

Question

Yasemin deposited  $1000 into a savings account. The relationship between the time,  t, in years, since the account was first opened, and Yasemin's account balance,  B(t), in dollars, is modeled by the following function.  B(t)=1000*e^(0.03 t) How many years will it take for Yasemin's account balance to reach  $1500 ? Round your answer, if necessary, to the nearest hundredth. years

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Given function and goal: We are given the function B(t) = 1000 * e^(0.03t) which models the account balance over time. We want to find the time t when the account balance B(t) reaches $1500. We can set up the equation 1500 = 1000 * e^(0.03t) and solve for t.Isolating exponential term: First, we divide both sides of the equation by 1000 to isolate the exponential term on one side. This gives us 1500 / 1000 = e^(0.03t).Simplifying equation: Simplifying the left side of the equation, we get 1.5 = e^(0.03t).Taking natural logarithm: To solve for t, we need to take the natural logarithm (ln) of both sides of the equation because the inverse of the exponential function is the natural logarithm. This gives us ln(1.5) = ln(e^(0.03t)).Solving for t: Using the property of logarithms that ln(e^x) = x, we can simplify the right side of the equation to 0.03t. So we have ln(1.5) = 0.03t.Calculating final value: To solve for t, we divide both sides of the equation by 0.03. This gives us t = ln(1.5) / 0.03.Now we calculate the value of ln(1.5) / 0.03 using a calculator. The calculation is t ≈ ln(1.5) / 0.03 ≈ 13.862943611198906 / 0.03 ≈ 13.86 years when rounded to the nearest hundredth.

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