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

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

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.

Thanks!! So basically that rule works for multiplication/division on both sides of the equation and not addition/subtraction?

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.

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

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

Thanks

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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