Find the time required for an investment of 5000 dollars to grow to 7700 dollars at an interest rate of 7.5 percent per year, compounded quarterly. Your answer is $$ t=\square $$ years. Question Help: Video Submit Question

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We start with the compound interest formula

$$
A = P \left(1 + \frac{r}{n}\right)^{n t},
$$

where  
$$ A $$ is the final amount,  
$$ P $$ is the principal,  
$$ r $$ is the annual interest rate (in decimal),  
$$ n $$ is the number of compounding periods per year, and  
$$ t $$ is the time in years.

Given:  
$$ A = 7700 $$,  
$$ P = 5000 $$,  
$$ r = 0.075 $$ (7.5%), and  
$$ n = 4 $$ (compounded quarterly).

Substitute the values into the formula:

$$
7700 = 5000 \left(1 + \frac{0.075}{4}\right)^{4t}.
$$

Simplify inside the parenthesis:

$$
1 + \frac{0.075}{4} = 1 + 0.01875 = 1.01875.
$$

Thus, the equation becomes:

$$
7700 = 5000 (1.01875)^{4t}.
$$

Divide both sides by 5000:

$$
\frac{7700}{5000} = (1.01875)^{4t}.
$$

Simplify the fraction:

$$
1.54 = (1.01875)^{4t}.
$$

Take the natural logarithm of both sides:

$$
\ln(1.54) = \ln\left((1.01875)^{4t}\right).
$$

Using the logarithm power rule, bring down the exponent:

$$
\ln(1.54) = 4t \cdot \ln(1.01875).
$$

Solve for $$ t $$:

$$
t = \frac{\ln(1.54)}{4 \ln(1.01875)}.
$$

Now, we calculate the logarithms:

$$
\ln(1.54) \approx 0.432,
$$
$$
\ln(1.01875) \approx 0.01856.
$$

Thus,

$$
t \approx \frac{0.432}{4 \times 0.01856} = \frac{0.432}{0.07424} \approx 5.82.
$$

So the time required is

$$
t \approx 5.82 \text{ years}.
$$

Therefore, your answer is $$ t \approx 5.82 $$ years.
The time required is approximately 5.82 years.

Find the time required for an investment of 5000 dollars to grow to 7700 dollars at an interest rate of 7.5 percent per year, compounded quarterly. Your answer is \( t=\square \) years. Question Help: Video Submit Question

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