Brain Teasers
Chemical Company
A chemical company spends x million dollars on research and finds that its profit can be expressed as a function of the amount of money spent on research. If Profit(x)= 30 + 6 log (x+2) how much will the company have to spend on research to increase its profit from its present level, with a research investment of 5 million dollars (P(5)), to 80 million dollars.
Hint
1) assume all number's have millions as units.Answer
P(5) = 30 + 6 log(5 + 2)----------------------------
log 2
= 46.84 million dollars
80 = 30 + 6 log2 (x + 2)
50 / 6 = log2 (x + 2)
250/6 = (x + 2)
x = 320.54 million dollars.
Therefore: increase in spending = 320.54 - 5 = 315.54 million
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This teaser's missing steps in the solution, and has at least one incorrect calculation with what's there. Furthermore, if the units of the formula are supposed to be in millions, that needs to be stated up front, not as a hint.
I must agree with paul726. There appears to be several problems with the solution. For instance, assuming 250/6 = (x+2), that would make (approximately) x = 39.7, not 320.54. How do you get from 50/6 = log(x+2) to 250/6 = x+2?
Okay. Total profit = P(x)=30+6log(x+2)
80=P(x+5)=30+6log(x+7)
Transitive property:
80=30+6log(x+7)
Subtraction:
50=6log(x+7)
Division:
25/3=log(x+7)
Now, if a=b then 10^a=10^b, right?
10^(25/3)=10^(log(x+7))
By the definition of logarithms:
10^(25/3)=x+7
Subtraction:
10^(25/3)-7=x
But 10^(25/3)-7 is 215443462, not 215.4. And my units were already in millions [I used P(x+5), not P(x+5000000)], so I really have 215.4 million million dollars to add to research!
80=P(x+5)=30+6log(x+7)
Transitive property:
80=30+6log(x+7)
Subtraction:
50=6log(x+7)
Division:
25/3=log(x+7)
Now, if a=b then 10^a=10^b, right?
10^(25/3)=10^(log(x+7))
By the definition of logarithms:
10^(25/3)=x+7
Subtraction:
10^(25/3)-7=x
But 10^(25/3)-7 is 215443462, not 215.4. And my units were already in millions [I used P(x+5), not P(x+5000000)], so I really have 215.4 million million dollars to add to research!
(Because there are no subscripts in the teaser text processor, it is necessary to use the arithmetic notation where "log2 j" is "binary logarithm of j" and "log10 k" is "common logarithm of k.")
The corrected teaser should include the following: If Profit(x)= 30 + 6 log2 (x+2) how much will the company have to spend on research to increase its profit from its present level, with a research investment of 5 million dollars (P(5)), to 80 million dollars.
Hint: 1) 30 and (x + 2) also have million dollars as units.
Answer: P(5) = 30 + 6 log2 (5 + 2) = 30 + 6 * (log10 (7) / log10 (2)) = 30 + 6 * (0.845 / 0.301) = 30 + 6 * 2.807 = 30 + 16.84 = 46.84
The initial profit is 46.84 million. Next, for an 80 million dollar profit from investment x.
80 = 30 + 6 log2 (x + 2)
50 / 6 = log2 (x + 2)
2^(50/6) = (x + 2)
2^(8.33333...) - 2 = x
x = 320.54
Therefore: increase in spending = 320.54 - 5 = 315.54 million.
The corrected teaser should include the following: If Profit(x)= 30 + 6 log2 (x+2) how much will the company have to spend on research to increase its profit from its present level, with a research investment of 5 million dollars (P(5)), to 80 million dollars.
Hint: 1) 30 and (x + 2) also have million dollars as units.
Answer: P(5) = 30 + 6 log2 (5 + 2) = 30 + 6 * (log10 (7) / log10 (2)) = 30 + 6 * (0.845 / 0.301) = 30 + 6 * 2.807 = 30 + 16.84 = 46.84
The initial profit is 46.84 million. Next, for an 80 million dollar profit from investment x.
80 = 30 + 6 log2 (x + 2)
50 / 6 = log2 (x + 2)
2^(50/6) = (x + 2)
2^(8.33333...) - 2 = x
x = 320.54
Therefore: increase in spending = 320.54 - 5 = 315.54 million.
Why is this still wrong?
The problem states log(x+2), not log2(x+2). The answer using log10(x+2) is completely impractical, so I kept looking for what I was doing wrong.
Apparently I'm not the first to notice the problem, but still it is here.
The problem states log(x+2), not log2(x+2). The answer using log10(x+2) is completely impractical, so I kept looking for what I was doing wrong.
Apparently I'm not the first to notice the problem, but still it is here.
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