Solve the equation $$ 4(1+r)^{15}=2 $$. Assume that $$ 1+r0 $$. Round answer to the nearest thousandth (3 decimal places). $$ r=\square $$

Question

Solve the equation $$ 4(1+r)^{15}=2 $$. Assume that $$ 1+r>0 $$. Round answer to the nearest thousandth (3 decimal places). $$ r=\square $$

UpStudy AI · Accepted Answer

1. Start with the equation:
   $$
   4(1+r)^{15}=2
   $$

2. Divide both sides of the equation by 4:
   $$
   (1+r)^{15}=\frac{2}{4}=\frac{1}{2}
   $$

3. Take the 15th root of both sides:
   $$
   1+r=\left(\frac{1}{2}\right)^{\frac{1}{15}}
   $$

4. Solve for $$ r $$:
   $$
   r=\left(\frac{1}{2}\right)^{\frac{1}{15}}-1
   $$

5. To compute the numerical value, express it using the exponential function:
   $$
   \left(\frac{1}{2}\right)^{\frac{1}{15}}=\exp\left(\frac{\ln\left(\frac{1}{2}\right)}{15}\right)=\exp\left(-\frac{\ln 2}{15}\right)
   $$
   Using $$\ln 2\approx0.693$$:
   $$
   \exp\left(-\frac{0.693}{15}\right)=\exp(-0.0462)\approx0.955
   $$

6. Substitute back to find $$ r $$:
   $$
   r\approx0.955-1=-0.045
   $$

Thus, the solution rounded to the nearest thousandth is:
$$
r\approx-0.045
$$
$$ r \approx -0.045 $$

Solve the equation . Assume that .
Round answer to the nearest thousandth (3 decimal places).

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Solve the equation . Assume that . Round answer to the nearest thousandth (3 decimal places).