Greg bought a gold coin for $9,000. If the value of the coin increases at a constant rate of 12% every 5 years, how many years will it take for the coin to be worth $20,000?
Approximately 35 years (35.23).
So we now want to find out when the value will be $20000. So
indicates that 20000/9000 = 20/9 = 1.12
Utilizing the logs from both sides, apply the LHS log rules to reduce the n:
The coin reaches $20000 value after roughly 35 years.
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To find out how many years it will take for the coin to be worth $20,000, we can use the formula for compound interest:
[ A = P \times (1 + r)^n ]
Where:
- ( A ) is the final amount (which is $20,000 in this case),
- ( P ) is the initial amount (which is $9,000),
- ( r ) is the interest rate (which is 12% or 0.12),
- ( n ) is the number of periods.
Rearranging the formula to solve for ( n ):
[ n = \frac{ \log \left( \frac{A}{P} \right) }{ \log(1 + r) } ]
Substituting the given values:
[ n = \frac{ \log \left( \frac{20,000}{9,000} \right) }{ \log(1 + 0.12) } ]
[ n ≈ \frac{ \log(2.222) }{ \log(1.12) } ]
[ n ≈ \frac{ 0.3465 }{ 0.0492 } ]
[ n ≈ 7.05 ]
So, it will take approximately 7 years for the coin to be worth $20,000.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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