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The temperature inside a certain industrial machine at time

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New post 17 Aug 2014, 09:31
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The temperature inside a certain industrial machine at time t seconds after startup, for 0 < t < 10, is given by h(t) = 4^(2t+1) – 4^(t+2) degrees Celsius. How many seconds after startup is the temperature inside the machine equal to 128 degrees Celsius?

(A) 3/2
(B) 2
(C) 5/2
(D) 3
(E) 7/2
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Re: The temperature inside a certain industrial machine at time  [#permalink]

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New post 17 Aug 2014, 09:42
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4^(2t+1) - 4^(t+2) = 128
=> (4^1)*(4^2t) - (4^2)*(4^t) = 128
=> 4*(4^t * 4^t) - 16*(4^t) = 128

let 4^t = x
=> 4(x^2) - 16x = 128
=> x^2 - 4x -32 =0
=> x=-4, 8

4^t cannot be negative for real x so
x=8
=> 4^t = 8
=> 2^2t = 2^3
=> t = 3/2

So, answer will be A

Hope it helps!
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New post 25 Nov 2014, 01:30
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for 0 < t < 10, is given by h(t) = 4^(2t+1) – 4^(t+2)

\(128 = 4^{2t+1} - 4^{t+2}\)

Dividing by 4

\(32 = 4^{2t} - 4^{t+1}\)

If t = 1, RHS = 0

If t = 2, RHS = \(4^4 - 4^3 = 256 - 64 = 192\)... This is way high than 32 (LHS)

So, 1 < t < 2 is the value of t

For \(t = \frac{3}{2}\), RHS = 64 - 32 = 32 = LHS

Answer = A
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Re: The temperature inside a certain industrial machine at time  [#permalink]

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New post 17 Aug 2014, 09:45
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oss198 wrote:
The temperature inside a certain industrial machine at time t seconds after startup, for 0 < t < 10, is given by h(t) = 4^(2t+1) – 4^(t+2) degrees Celsius. How many seconds after startup is the temperature inside the machine equal to 128 degrees Celsius?

(A) 3/2
(B) 2
(C) 5/2
(D) 3
(E) 7/2


We are asked to find t for which h(t) equals to 128, so to solve for t, the following equation: 128 = 4^(2t+1) – 4^(t+2).

Personally, I would plug answers for this question.

A works: 4^(2*3/2+1) – 4^(3/2+2) = 4^4 - 4^(7/2) = 2^8 - 2^7 = 2^7 = 128.

Answer: A.
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Re: The temperature inside a certain industrial machine at time t seconds  [#permalink]

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New post 24 Nov 2014, 05:09
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Hi,
So I see how to solve this from MGMT but I want to know what led me astray in my own solution. Not sure what rule I'm mixing up. Here's how I approached it (incorrect)

4^(2t+1) - 4^(t+2) = 128
4^(2t+1) - 4^(t+2) = 4^3+4^3

therefore
2t+1-(t+2) = 3+3
2t+1-t-2 = 6
t-1= 6
t= 7


So am I incorrect in saying those powers equal? I thought you could do that if the bases were the same. What am I missing please.
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Re: The temperature inside a certain industrial machine at time  [#permalink]

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New post 24 Nov 2014, 10:24
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HI,

You can equate the powers on both the sides if the base is same IS CORRECT logic, but you applied it wrong.
4^(2t+1) - 4^(t+2) = 4^3+4^3
this CANNOT be equated as the base is not same. To throw more light on this:--
2^a * 3^b = 2^20 * 3^30
so, a= 20 and b=30
We can equate this as we have broken down both the sides into product of powers of prime numbers. But in
4^(2t+1) - 4^(t+2) = 4^3+4^3

4^(2t+1) and 4^(t+ 2) are subtracted.. they are not expressed as "product of powers of prime numbers". Same is the case with right hand side.. so we cannot equate the powers directly!
(Check my solution as the reply to this previous post for more clarity)

Hope it helps!

angelfire213 wrote:
Hi,
So I see how to solve this from MGMT but I want to know what led me astray in my own solution. Not sure what rule I'm mixing up. Here's how I approached it (incorrect)

4^(2t+1) - 4^(t+2) = 128
4^(2t+1) - 4^(t+2) = 4^3+4^3

therefore
2t+1-(t+2) = 3+3
2t+1-t-2 = 6
t-1= 6
t= 7


So am I incorrect in saying those powers equal? I thought you could do that if the bases were the same. What am I missing please.

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Re: The temperature inside a certain industrial machine at time  [#permalink]

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New post 24 Nov 2014, 12:00
Thanks!! So basically that rule works for multiplication/division on both sides of the equation and not addition/subtraction?
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New post 25 Nov 2014, 01:35
angelfire213 wrote:
Thanks!! So basically that rule works for multiplication/division on both sides of the equation and not addition/subtraction?



That's correct

Example:

\(4^a * 4^b = 4^{(a+b)}\)

\(4^a + 4^b = 4^a (1 + 4^{(b-a)})\)
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Re: The temperature inside a certain industrial machine at time  [#permalink]

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New post 12 Mar 2015, 03:22
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Bunuel wrote:
oss198 wrote:
The temperature inside a certain industrial machine at time t seconds after startup, for 0 < t < 10, is given by h(t) = 4^(2t+1) – 4^(t+2) degrees Celsius. How many seconds after startup is the temperature inside the machine equal to 128 degrees Celsius?

(A) 3/2
(B) 2
(C) 5/2
(D) 3
(E) 7/2


We are asked to find t for which h(t) equals to 128, so to solve for t, the following equation: 128 = 4^(2t+1) – 4^(t+2).

Personally, I would plug answers for this question.

A works: 4^(2*3/2+1) – 4^(3/2+2) = 4^4 - 4^(7/2) = 2^8 - 2^7 = 2^7 = 128.

Answer: A.



Should we expect such question on real GMAT , it seems we have to carefully choose the seed from the answer choices to start with and then go backwards or forward in this case.
though it is an easy one but we cannot deny it takes some time to calculate numbers with pen on paper.
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Re: The temperature inside a certain industrial machine at time  [#permalink]

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New post 10 Apr 2015, 07:19
oss198 wrote:
The temperature inside a certain industrial machine at time t seconds after startup, for 0 < t < 10, is given by h(t) = 4^(2t+1) – 4^(t+2) degrees Celsius. How many seconds after startup is the temperature inside the machine equal to 128 degrees Celsius?

(A) 3/2
(B) 2
(C) 5/2
(D) 3
(E) 7/2


at t = 1, temp would be 0 deg
at t = 2, temp would be > 128 deg
so 128 deg would be between 1 and 2.
only option A fits.

otherwise solve it making a quadratic or substitute options.
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Re: The temperature inside a certain industrial machine at time  [#permalink]

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New post 05 Jan 2016, 04:11
oss198 wrote:
The temperature inside a certain industrial machine at time t seconds after startup, for 0 < t < 10, is given by h(t) = 4^(2t+1) – 4^(t+2) degrees Celsius. How many seconds after startup is the temperature inside the machine equal to 128 degrees Celsius?

(A) 3/2
(B) 2
(C) 5/2
(D) 3
(E) 7/2


I didn't make any complex manipulations, I've just started with asnwer choice B.
\(4^5-4^4=4^4*(4-1)=4^4*3=128?\) This number \(4^4*3\) is already too big and > 128 so we need a smaller. The only smaller number among the answer choices is 3/2.
Answer A
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Re: The temperature inside a certain industrial machine at time  [#permalink]

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New post 25 Jun 2016, 00:34
Hi a question on this: why can't we convert it into powers of base 2?

so 128 = 2^7
and so we have 2^7= 2^2(2t+1) - 2^2(t+2)?
then we can equate 7= 2(2t+1)- 2(t+2)?

Can someone please explain why this is wrong?

Thank you!
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New post 25 Jun 2016, 02:47
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Re: The temperature inside a certain industrial machine at time  [#permalink]

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New post 26 Jun 2016, 01:51
h = 4^(2t+1) – 4^(t+2) = 2^(4t+2) - 2^(2t+4)

We observe that h increases pretty fast with t. If t=2 for example, the first expression becomes 1,024, and the second expression is 256. So t has to be pretty small. 3/2 is the smallest value and satisfies the condition, so the correct answer choice is A.
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Re: The temperature inside a certain industrial machine at time  [#permalink]

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New post 27 Mar 2017, 13:48
Bunuel wrote:
raashibanka wrote:
Hi a question on this: why can't we convert it into powers of base 2?

so 128 = 2^7
and so we have 2^7= 2^2(2t+1) - 2^2(t+2)?
then we can equate 7= 2(2t+1)- 2(t+2)?

Can someone please explain why this is wrong?

Thank you!


This is mathematically wrong: \(2^3=2^4-2^3\) but \(3\neq 4-3\).


Thanks Bunuel! What if made 4 the common base instead:

so 128 = 2*64 = 2*4^3 = 4^1/2*4^3=4^3/2 which happens to be the correct answer choice.

But - unless I made a math mistake - doesn't seem to work algebraically if we get rid of the bases and set the exponents equal to one another

3/2 = (2t -1) - (t-2) = 2t-1-t+2 => 3/2 = t+1 => t=1/2

Where am going wrong here? And is just a coincidence that 128 simplifies to 4^3/2 (assuming my math is right there)? I'm slow with testing cases, so an alternative approach to these problems would be awesome! :-D

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Re: The temperature inside a certain industrial machine at time  [#permalink]

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New post 28 Mar 2017, 03:29
dreidco wrote:
Bunuel wrote:
raashibanka wrote:
Hi a question on this: why can't we convert it into powers of base 2?

so 128 = 2^7
and so we have 2^7= 2^2(2t+1) - 2^2(t+2)?
then we can equate 7= 2(2t+1)- 2(t+2)?

Can someone please explain why this is wrong?

Thank you!


This is mathematically wrong: \(2^3=2^4-2^3\) but \(3\neq 4-3\).


Thanks Bunuel! What if made 4 the common base instead:

so 128 = 2*64 = 2*4^3 = 4^1/2*4^3=4^3/2 which happens to be the correct answer choice.

But - unless I made a math mistake - doesn't seem to work algebraically if we get rid of the bases and set the exponents equal to one another

3/2 = (2t -1) - (t-2) = 2t-1-t+2 => 3/2 = t+1 => t=1/2

Where am going wrong here? And is just a coincidence that 128 simplifies to 4^3/2 (assuming my math is right there)? I'm slow with testing cases, so an alternative approach to these problems would be awesome! :-D

Thanks


Not sure what you are doing there but \(\frac{4^1}{2}*4^3 = \frac{4^4}{2}\), not \(\frac{4^3}{2}\). Also, the answer is 3/2. You got there (incorrectly) \(\frac{4^3}{2}\), how is this the answer?
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Re: The temperature inside a certain industrial machine at time  [#permalink]

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New post 30 Mar 2017, 16:42
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oss198 wrote:
The temperature inside a certain industrial machine at time t seconds after startup, for 0 < t < 10, is given by h(t) = 4^(2t+1) – 4^(t+2) degrees Celsius. How many seconds after startup is the temperature inside the machine equal to 128 degrees Celsius?

(A) 3/2
(B) 2
(C) 5/2
(D) 3
(E) 7/2


We are given that the temperature inside a certain industrial machine at time t seconds after startup, for 0 < t < 10, is given by h(t) = 4^(2t+1) – 4^(t+2) degrees Celsius. We need to determine how many seconds after startup it takes the temperature inside the machine to equal 128 degrees Celsius. Thus:

4^(2t+1) – 4^(t+2) = 128

(4^2t)(4^1) - (4^t)(4^2) = 128

4(4^2t) - 16(4^t) - 128 = 0

If we let x = 4^t, the above equation can be turned into:

4x^2 - 16x - 128 = 0

x^2 - 4x - 32 = 0

(x - 8)(x + 4) = 0

x = 8 or x = -4

Since x = 4^t, we have 4^t = 8 or 4^t = -4. However, since 4 is positive, any power it raises to will be positive also, so 4^t can’t be equal to -4. Thus, we only need to solve 4^t = 8:

4^t = 8

(2^2)^t = 2^3

2^2t = 2^3

2t = 3

t = 3/2

Answer: A
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Re: The temperature inside a certain industrial machine at time  [#permalink]

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New post 04 Jun 2017, 19:19
oss198 wrote:
The temperature inside a certain industrial machine at time t seconds after startup, for 0 < t < 10, is given by h(t) = 4^(2t+1) – 4^(t+2) degrees Celsius. How many seconds after startup is the temperature inside the machine equal to 128 degrees Celsius?

(A) 3/2
(B) 2
(C) 5/2
(D) 3
(E) 7/2


This problem should be backsolved. We have relatively user-friendly answer choices, so we can start with (B).

(B)
4^(2*2+1) - 4^(4) = 4^5 - 4^4 = 4^4(4-1) = 4^4(3)

4^4 = 2^8 = 256.

256*3 > 128, so we already can eliminate all other answer choices and only check (A) for sanity's sake.

(A) 4^(2*3/2+1) - 4^(3/2+2)

4^4 - 4^7/2

2^8 - 2^7 = 256 - 128 = 128.
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