Answer:
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A)
to solve a, the following rules are crucial:
(i)
so the difference of 2 logarithms with the same base, is the logarithm of their division, preserving the same base.
(ii) if
then b=c.
so if 2 logarithms with the same base are equal, then the arguments (b and c) are equal as well.
so
apply rule (i):
apply rule (ii):
now 'isolate' y:
b)
some more rules:
(iii)
(iv)
apply iii and iv:
then substitute
in the equation:
so now we have a quadratic equation of degree 2,
a=2, b=-5, c=2
the discriminant is
, the root of it is 3
so the roots are:
and
finally, we convert u's to x's:
means
so for u=2, x=4, and for u=1/2 we have x=1/4
Answers:
A) y=(6 x^{2} -1) \frac{(x-1)}{x}
B) solution set: {4, 1/4}
Given that the printer depreciates at the rate of 14% p.a. This has been modeled by the function V=2400(1-0.14)^t, this follows an exponential form given by
y=a(b)^x
where:
V=y
a=2400
b=1-0.14
x=t
thus:
<span>Part A: Explain what the parameter 2,400 represents in the equation of the function.
</span>The parameter 2400 represents the initial value of the printer at time t=0. This is the original value.
<span>Part B: What is the factor by which the printer depreciates each year?
The factor of depreciation is 14% percent. This is the rate at which the printer depreciates and it accounts for the value of the printer at the end of every year.
</span><span>Part C: Amy also considered purchasing a printer that costs $4,000 and depreciates by 25% each year. Which printer will have more value in 5 years?
Value after 5 years of the $2400 printer that depreciates at 14% per year will be:
V(t)=2400(1-0.14)^5=$1,129.025
Value after 5 years of the $4000 printer that depreciates at 25% per year will be:
F(t)=4000(1-0.25)^t
F(5)=4000(1-0.25)^5=$949.22
The printer that costs $2400 will be more valuable compared to the printer that cost $4000</span>