The graph below shows the estimated number of AI-optimized semiconduct

Question

The graph below shows the estimated number of AI-optimized semiconductor chips (in millions) produced globally each year from 2014 to 2024. As AI applications expand across industries, production has followed an exponential growth pattern, modeled by the equation:

\(y = P \cdot 2^{k(t - 2014)}\)

where:

y is the number of chips produced (in millions),
t is the year,
P and k are positive constants.

From each drop-down menu, select the option that best completes the statement based on the information provided.

The value of the constant P is approximately and if the model remains accurate, the number of chips produced will be approximately double the 2024 value in the year .

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Official Answer

Dropdown 1: 10 Dropdown 2: 2027

APram · Answer

For value of P

Substitute at t=2014, y=10
=> P × 2^0 = 10
=> P = 10

Now at t=2024, y=150
=> 150 = 10×(2^10k)
=> 15 = 2^10k

Approximating 16=2^10k
=> 10k ~ 4
=> k = 0.4

So for double in t years

10×2^0.4(t-2014) = 10×(2^4)×2 = 10×2^5
=> 0.4t = 805.6+5 = 810.6
=> t = 2026.5

We will have 2027

So answer is  10 and 2027

Phoenix2107 · Answer

Hello  APram ,

Can u explain the third step?

I understood how we arrived at P and k values ..can u explain how u arrived in getting t?

APram · Answer

We can see that at t = 2024, y = 150

So Substituting the values we get the equation

150 = 10 × 2^10k

Here to arrive value of k easily, I approximated 15 as 16
So 2^4 = 2^10k
So approximately 10k = 4
Hence k = 0.4

Now we are asked that the value of y will get doubled than 2024 at t = ?
So doubling value of y we get

2 × 2^10k = 2^(10k+1)
Also as per equation this is 2^(k×(t-2014))

Equating both
We get 10k+1 = k(t-2014)
Apply the value of k=0.4
We get t= 2026.5
Since we had approximated value of k earlier, so we should take care of that
Since we had taken more than required, so t must be more than 2026
Hence 2027.

Vibhatu · Answer

need to do all this complex calculation. put what value of 2^ get 30, we can say 2^5. since k=0.4, just put it we will get 12.5 thus year must me 2026.5. answer is 2027

velvetwhisper · Answer

To find value of P,

Substitute t= 2014 & y= 10
=> 10 = P x 2^k(2014- 2014)
=> P x 2^0 = 10
=> P = 10

So answer to Dropdown 1 is 10

For T= 2024, y= 150
So 150 = 10 x 2^k(2024- 2014)
=> 10 x 2^10k = 150
=> 2^10k = 15
(To get the value of “k” easily, I took 15 as 16 which is equals to 2^4)
So 2^10k = 16= 2^4
=> k = 0.4

Since the question says the number of chips produced will be approximately double the 2024 value, so “y” will be

150 x 2 = 300 (value of y in 2024 is 150 as per graph)

Now replacing the above values we have so far, we get

300= 10 x 2^k(t- 2014)
=> 30 = 2^0.4 (t-2014)
=> 32 = 2^0.4 (t-2014)  
=> 2^5 = 2^0.4 (t-2014)
=> 5 = 0.4(t-2014)
=> t- 2014 = 12.5
=> t= 2026.5 ~ 2027

So answer to Dropdown 2 is 2027

agrasan · Answer

To get the value of P
put t = 2014 and y respectively is 10 which means that P = 10

To get the value of k, 
we can see that value of y is ~20 at t = 2017
from the equation, we get  20 = 10*(2^k(2017-2014)) 
solving k is easy from here by equating exponents k ~ 1/3

Now the question becomes very simple, we just have to find when value becomes double that of in 2024 which means that (2^k(t-2014)) should be 2, can only happen with t = 2027.

Bunuel · Answer

Solution: 
From the equation  y = P \cdot 2^{k(t - 2014)} , when  t = 2014 , the exponent becomes zero and  y = P . According to the graph, production in 2014 was approximately 10 million chips, so  P = 10 .
To find when the production doubles the 2024 value (~150 million), note that the model shows exponential growth with doubling every 2.5 years. Therefore, production would reach ~300 million around  2024 + 2.5 = 2026.5 , which rounds to  2027 .

Answers: 
 
 Constant  P :  10  
 Year when production doubles from 2024 level:  2027

The graph below shows the estimated number of AI-optimized semiconduct

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The graph below shows the estimated number of AI-optimized semiconduct

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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