To find when the two savings accounts will have the same balance, we need to set the two equations equal to each other:
Bank #1:
\[ y = 12,150(1.023)^x \]
Bank #2:
\[ y = 12,025(1.029)^x \]
We set them equal to find \( x \):
\[ 12,150(1.023)^x = 12,025(1.029)^x \]
To solve for \( x \), we can divide both sides by \( 12,025 \):
\[ \frac{12,150}{12,025}(1.023)^x = (1.029)^x \]
Next, calculate \( \frac{12,150}{12,025} \):
\[ \frac{12,150}{12,025} \approx 1.0104 \]
Now the equation looks like this:
\[ 1.0104(1.023)^x = (1.029)^x \]
Now take the natural logarithm of both sides:
\[ \ln(1.0104) + x \ln(1.023) = x \ln(1.029) \]
Now we can isolate \( x \):
\[ \ln(1.0104) = x(\ln(1.029) - \ln(1.023)) \]
Now, calculate \( \ln(1.0104) \) and \( \ln(1.029) - \ln(1.023) \):
- \( \ln(1.0104) \approx 0.0104 \)
- \( \ln(1.029) \approx 0.0285 \)
- \( \ln(1.023) \approx 0.0227 \)
- Thus, \( \ln(1.029) - \ln(1.023) \approx 0.0285 - 0.0227 = 0.0058 \)
So now we have:
\[ 0.0104 = x(0.0058) \]
\[ x \approx \frac{0.0104}{0.0058} \approx 1.7931 \]
Rounding to the nearest tenth, we find that:
The two accounts will have the same balance in about 1.8 years.
To determine which account will yield more money in the long run, we can analyze the limits of the two equations as \( x \) approaches infinity:
The account from Bank #1 grows at a rate of \( 1.023 \) and from Bank #2 at \( 1.029 \). Since \( 1.029 > 1.023 \), Bank #2 will yield more money in the long run.
So, the final answers are:
The two accounts will have the same balance in about 1.8 years. In the long run, the account from bank #2 will yield more money.