Math  /  Algebra

QuestionSection 5.2 Problem 31 An investment initially worth $5300\$ 5300 earns 7.5%7.5 \% annual interest, and an investment initially worth $7900\$ 7900 earns 5.8%5.8 \% annual interest, both compounded annually.
How long will it take for the smaller investment to catch up with the larger one?
It will take \square years.

Studdy Solution
Solve the equation for the number of years t t .
Divide both sides by 5300 to isolate the exponential terms:
(1.075)t=79005300×(1.058)t (1.075)^t = \frac{7900}{5300} \times (1.058)^t
Simplify the fraction:
(1.075)t=7953×(1.058)t (1.075)^t = \frac{79}{53} \times (1.058)^t
Take the natural logarithm of both sides:
ln((1.075)t)=ln(7953×(1.058)t) \ln((1.075)^t) = \ln\left(\frac{79}{53} \times (1.058)^t\right)
Use the properties of logarithms to separate terms:
tln(1.075)=ln(7953)+tln(1.058) t \cdot \ln(1.075) = \ln\left(\frac{79}{53}\right) + t \cdot \ln(1.058)
Rearrange to solve for t t :
tln(1.075)tln(1.058)=ln(7953) t \cdot \ln(1.075) - t \cdot \ln(1.058) = \ln\left(\frac{79}{53}\right)
Factor out t t :
t(ln(1.075)ln(1.058))=ln(7953) t (\ln(1.075) - \ln(1.058)) = \ln\left(\frac{79}{53}\right)
Solve for t t :
t=ln(7953)ln(1.075)ln(1.058) t = \frac{\ln\left(\frac{79}{53}\right)}{\ln(1.075) - \ln(1.058)}
Calculate t t :
tln(1.490566)ln(1.075)ln(1.058) t \approx \frac{\ln(1.490566)}{\ln(1.075) - \ln(1.058)}
t0.3990.017 t \approx \frac{0.399}{0.017}
t23.47 t \approx 23.47
Since t t must be a whole number, round up to the nearest whole number:
It will take 24 \boxed{24} years.

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