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4 years ago

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The Mendes family bought a new house 8 years ago $129,000. The new house is now worth $186,000. Assuming a steady rate of growth

, what was the yearly rate of appreciation? Round your answer to the nearest tenth of a percent (1.2% etc)
******PLEASE HELP****

1 answer:

USPshnik [31]4 years ago

3 0

So assuming simplet interest

find the change to find the interest
change=186000-129000=57000

simple interest=time times rate times principal
time is in years (8)
rate is in decimal
principal is amount invested (129000)

so
interest is 57000
57000=(8)(r)(129000)
57000=1032000r
divide both sides by 1032000
0.05523255813953488372093023255814=r
or
5.523255813953488372093023255814%
rounded
5.5%

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Step-by-step explanation:

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Step-by-step explanation:

<u><em>Explanation</em></u> :-

Given logarithmic function $y = log_{b} (x)$ if b >1

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<em>put x = </em> $\frac{1}{4}$ <em> given b > 1 so we can choose b = 2</em>

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<em>iii) </em>

<em>put x = </em> $\frac{1}{2}$ <em> given b > 1 so we can choose b = 2</em>

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<em>we will apply logarithmic formula </em>

<em>log x ⁿ = n log (x)</em>

$y = log_{2} (2^{-1} ) = -1 log_{2} (2) = - (1) = -1$

<em>y = -1</em>

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