Solve the given problem involving logarithms

In summary: S_n=\dfrac{a(r^n -1)}{r-1}## our lack of agreement seems to be on the semantics as to whether the first term is ab or a.In summary,The linear function, \log S, increases linearly with time.Online business may be affected with different conditions like taxation in other countries that may in turn slow down the sales. Sales may fluctuate up and down and the model may not have considered that aspect...as it projects continous growth in sale numbers and no decline.These things may all be true, but they are not the right answer.Still, it's a long way to extrapolate from
  • #1
chwala
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Homework Statement
See attached.
Relevant Equations
understanding of logs
1686712212308.png
Part (a)

##s=ab^t##

##\log s= \log a+ t\log b##

Expression on the right hand side increases linearly with ##t##Part(b)

1686712380323.png


##s=120 ×1.15^t##

##\log s = \log 120+t\log 1.15##

##\log s =2.08+0.06t##

From graph, y-intercept = ##2.08##

##m=\dfrac{2.45-2.08}{6-0}=0.06##

Part (c)

##s(7)=319.2##

##s(6)=277.56##

##s_7Required = 319.2-277.56=41.6##

##s_7Required = 319.2-277.56=42##

Part (d)i

##120 ×1.15^t>70,000##

##1.15^t>583.3##

##t\log 1.15>\log 583.3##

...

##t=46##

Part (d)ii
Online business may be affected with different conditions like taxation in other countries that may in turn slow down the sales. Sales may fluctuate up and down and the model may not have considered that aspect...as it projects continous growth in sale numbers and no decline.Your insights...thanks guys!
 
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  • #2
You can see (assuming your line is accurate) that as [itex]t[/itex] increases, the observations fall visibly further and further below the line predicted by the model. This suggests that extrapolation to [itex]t \approx 46[/itex] from [itex]t = 5[/itex] and [itex]t = 6[/itex] is likely to give an overestimate.

Reasons as to why the business will not sustain exponential sales growth over the next three years belong to the realm of economics, rather than basic mathematics, but an obvious one is that there is an upper limit to the number of products a small business can produce in a month.
 
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  • #3
I think the line is drawn inaccurately. The points seem to fall exactly on the line 2.08 + 0.06t. Still, it's a long way to extrapolate from t = 6 to 46...

More seriously, I think there's a misinterpretation. S is the number of items sold in a month, not the cumulative total sales. So the number sold in the seventh month is S(7), not S(7) - S(6). For part d, you will need to get an expression for the cumulative total.
 
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  • #4
mjc123 said:
I think the line is drawn inaccurately. The points seem to fall exactly on the line 2.08 + 0.06t. Still, it's a long way to extrapolate from t = 6 to 46...

More seriously, I think there's a misinterpretation. S is the number of items sold in a month, not the cumulative total sales. So the number sold in the seventh month is S(7), not S(7) - S(6). For part d, you will need to get an expression for the cumulative total.
I hear you...maybe, I should think along the lines of finding the area bound by the curve between ##t=0## and ##t=46## using integration...thinking along these lines...our function being a straight line.

...or use sum of A.P because the function is linear.
 
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  • #5
chwala said:
I hear you...maybe, I should think along the lines of finding the area bound by the curve between ##t=0## and ##t=46## using integration...thinking along these lines...our function being a straight line.

...or use sum of A.P because the function is linear.

[itex]\log S[/itex] is linear, but you want [tex]\sum_{n=1}^N S(n) = a \sum_{n=1}^N b^n.[/tex]
 
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  • #6
chwala said:
Part (d)ii
Online business may be affected with different conditions like taxation in other countries that may in turn slow down the sales. Sales may fluctuate up and down and the model may not have considered that aspect...as it projects continous growth in sale numbers and no decline.
These things may all be true, but they are not the right answer.
mjc123 said:
Still, it's a long way to extrapolate from t = 6 to 46...
Is along the right lines (although t is not 46, and it's not the increase in t that is the problem).
 
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  • #7
chwala said:
I hear you...maybe, I should think along the lines of finding the area bound by the curve between ##t=0## and ##t=46## using integration...thinking along these lines...our function being a straight line.

...or use sum of A.P because the function is linear.
Sum of GP I would say. ##T_1=10^{2.14}## or ##T_1=120 \times 1.15## and ##r=1.15##
 
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  • #8
neilparker62 said:
Sum of GP I would say. ##T_1=10^{2.14}## or ##T_1=120 \times 1.15## and ##r=1.15##
...Am getting

##\dfrac{138(1.15^t -1)}{0.15} >70,000##

##1.15^t >77.08##

##t>31##

...i hope this is correct...cheers guys!
 
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  • #9
chwala said:
##\dfrac{138(1.15^t -1)}{0.15} >70,000##
Where does 138 come from?
 
  • #10
pbuk said:
Where does 138 come from?
##a=138##.
 
  • #11
## a = 120 \ne 138 ##
 
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  • #12
pbuk said:
## a = 120 \ne 138 ##
Let me look at it again. Cheers @pbuk
 
  • #13
chwala said:
##t>31##

...i hope this is correct...cheers guys!
FWIW, using a short (5 lines of BASIC) program I found 70,000 is first exceeded during month 32.
 
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  • #14
138 = ab = S1, the first term in the sum.
Total = ab(bn - 1)/(b-1)
I have said before that you need to show more of your working, explain where your numerical expressions come from.
 
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  • #15
mjc123 said:
138 = ab = S1, the first term in the sum.
Total = ab(bn - 1)/(b-1)
I have said before that you need to show more of your working, explain where your numerical expressions come from.
My working was just fine! what you are calling ab is what i am calling a. Most importantly, i used the Sum of GP as required. That is,

##S_n=\dfrac{a(r^n -1)}{r-1}## our lack of agreement seems to be on the semantics as to whether the first term is ab or a!
 
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  • #16
chwala said:
Let me look at it again. Cheers @pbuk
##a=138≠120##.

The terms in the series are as follows,

##[(120 ×1.15^1) + (120 ×1.15^2) +(120 ×1.15^3)+ ...)##

More specifically,

##[138 +158.7+182.505 + ...)##
 
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  • #18
Steve4Physics said:
FWIW, using a short (5 lines of BASIC) program I found 70,000 is first exceeded during month 32.
As per OP's calculation in post #8.
 
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  • #19
chwala said:
My working was just fine!
No, your result was correct but you didn't show your workings.

The point is that it is easy to make mistakes when using the equalities
$$ \sum_{k=0}^{n-1} ar^k
= \sum_{k=1}^{n} ar^{k-1}
= a\left(\frac{1-r^{n}}{1-r}\right)
= a\left(\frac{r^{n}-1}{r-1}\right) $$
because of the ## n-1 / k -1 ## thing. You need to show that your choice of ## a, r \text{ and } n ## match the terms of the question so the first term is ## ar^0 ## and the last term is ## ar^{(n-1)} ##, otherwise it is easy to use the wrong ## a ## or to get ## n ## wrong by 1.
 
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