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Helen [10]
2 years ago
13

Write an expression for the sequence of operations described below.

Mathematics
1 answer:
NeX [460]2 years ago
3 0

Answer:

(7+z) - y

Step-by-step explanation:

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Math help questions <br>​
Evgen [1.6K]

After 1 year, the initial investment increases by 7%, i.e. multiplied by 1.07. So after 1 year the investment has a value of $800 × 1.07 = $856.

After another year, that amount increases again by 7% to $856 × 1.07 = $915.92.

And so on. After t years, the investment would have a value of \$800 \times 1.07^t.

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856 \times 1.07 ^n = 1400

1.07^n = \dfrac74

\log_{1.07}\left(1.07^n\right) = \log_{1.07}\left(\dfrac74\right)

n \log_{1.07}(1.07) = \log_{1.07} \left(\dfrac74\right)

n = \log_{1.07} \left(\dfrac74\right) = \dfrac{\ln\left(\frac74\right)}{\ln(1.04)} \approx \boxed{8.3}

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