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3 years ago

9

The expression 1000(1.1)^t represents the value of a $1000 investment that earns 10% interest per year, compounded annually for

t years. What is the value of a $1000 investment at the end of 4 years? Do not use a unit label or comma when enter the answer

2 answers:

torisob [31]3 years ago

8 0

After 4 years
the value = 1000 × (1.1)⁴
the value = 1000 × 1.4641
the value = 1464.1

kakasveta [241]3 years ago

4 0

Answer: 1464.1

Step-by-step explanation:

Given : The expression $1000(1.1)^t$ represents the value of a $1000 investment that earns 10% interest per year, compounded annually for t years.

To find the value of a $1000 investment at the end of 4 years, we substityte x = 4 in the given expression , then

$1000(1.1)^4=1464.1$

Therefore, the value of a $1000 investment at the end of 4 years = 1464.1

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Step-by-step explanation:

Let as consider the given equations are $\log_3(\dfrac{81}{3})=?,\log_5(\dfrac{625}{25})=?,\log_2(\dfrac{64}{8})=?,\log_4(\dfrac{64}{16})=?,\log_6(36^4)=?,\log(100^3)=?$ .

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(c)

$\log_2(\dfrac{64}{8})=\log_2(8)$

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$\log_2(\dfrac{64}{8})=3$ $[\because \log_aa^x=x]$

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$\log_4(\dfrac{64}{16})=1$ $[\because \log_aa^x=x]$

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$\log_6(36^4)=\log_6((6^2)^4)$

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$\log_6(36^4)=8$ $[\because \log_aa^x=x]$

(f)

$\log(100^3)=\log((10^2)^3)$

$\log(100^3)=\log(10^6)$

$\log(100^3)=6$ $[\because \log10^x=x]$

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