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Tammy deposits [tex]$\$[/tex]1,850[tex]$ in an individual retirement account earning $[/tex]2.6\%[tex]$ interest, compounded annually. She also deposits $[/tex]\[tex]$2,015$[/tex] in a business interest-bearing account earning [tex]$1.5\%$[/tex] interest, compounded annually.

Select the equation and the number of years, [tex]$x$[/tex], it will take for the amount of money in both accounts to be equal. Round to the nearest whole year.

A. [tex]$1,850(1.026)^x = 2,015(1.015)^x$[/tex]
B. 9 years
C. 6 years
D. [tex]$1,850(1.126)^x = 2,015(1.115)^x$[/tex]
E. [tex]$1,850(1.26)^x = 2,015(1.15)^x$[/tex]
F. 8 years



Answer :

GinnyAnswer
To determine the number of years it takes for the amounts in both accounts to be equal, we need to set up an equation for the amount of money in each account after [tex]\( x \)[/tex] years. Then we will set these equations equal to each other and solve for [tex]\( x \)[/tex].

Step-by-step solution:

1. Calculate the amount in the retirement account after [tex]\( x \)[/tex] years:
- Principal [tex]\( P_1 = \$1850 \)[/tex]
- Interest rate [tex]\( r_1 = 2.6\% = 0.026 \)[/tex]
- Formula for compound interest: [tex]\( P_1(1 + r_1)^x \)[/tex]

So, the amount in the retirement account after [tex]\( x \)[/tex] years is:
[tex]\[ 1850(1 + 0.026)^x = 1850(1.026)^x \][/tex]

2. Calculate the amount in the business account after [tex]\( x \)[/tex] years:
- Principal [tex]\( P_2 = \$2015 \)[/tex]
- Interest rate [tex]\( r_2 = 1.5\% = 0.015 \)[/tex]
- Formula for compound interest: [tex]\( P_2(1 + r_2)^x \)[/tex]

So, the amount in the business account after [tex]\( x \)[/tex] years is:
[tex]\[ 2015(1 + 0.015)^x = 2015(1.015)^x \][/tex]

3. Set the equations equal to each other to find [tex]\( x \)[/tex]:
[tex]\[ 1850(1.026)^x = 2015(1.015)^x \][/tex]

4. Solve for [tex]\( x \)[/tex]:
Divide both sides by 1850:
[tex]\[ (1.026)^x = \frac{2015}{1850} (1.015)^x \][/tex]

Simplify the fraction:
[tex]\[ (1.026)^x = 1.089189189 (1.015)^x \][/tex]

Take the natural logarithm of both sides:
[tex]\[ \ln((1.026)^x) = \ln(1.089189189) + \ln((1.015)^x) \][/tex]

Use the property of logarithms:
[tex]\[ x \ln(1.026) = \ln(1.089189189) + x \ln(1.015) \][/tex]

Rearrange to solve for [tex]\( x \)[/tex]:
[tex]\[ x \ln(1.026) - x \ln(1.015) = \ln(1.089189189) \][/tex]

Factor out [tex]\( x \)[/tex]:
[tex]\[ x (\ln(1.026) - \ln(1.015)) = \ln(1.089189189) \][/tex]

Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{\ln(1.089189189)}{\ln(1.026) - \ln(1.015)} \][/tex]

5. Calculate the logarithms:
- [tex]\(\ln(1.089189189) \approx 0.085307\)[/tex]
- [tex]\(\ln(1.026) \approx 0.02573\)[/tex]
- [tex]\(\ln(1.015) \approx 0.014925\)[/tex]

Therefore:
[tex]\[ x \approx \frac{0.085307}{0.02573 - 0.014925} \approx \frac{0.085307}{0.010805} \approx 7.895 \][/tex]

6. Round [tex]\( x \)[/tex] to the nearest whole year:
[tex]\[ x \approx 8 \][/tex]

Summary:

- The correct equation is [tex]\( 1850(1.026)^x = 2015(1.015)^x \)[/tex]
- The rounded number of years [tex]\( x \)[/tex] is [tex]\( 8 \)[/tex] years.

So, the correct answers are:
- [tex]\( 1850(1.026)^2=2,015(1.015)^x \)[/tex]
- 8 years

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