The temperature inside a certain industrial machine at time

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.

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.

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!

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

That's correct

Example:

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

\(4^a + 4^b = 4^a (1 + 4^{(b-a)})\)

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.

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.

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

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 \(3\neq 4-3\).

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.

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

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?

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

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.

BrentGMATPrepNow VeritasKarishma

pls explain what`s wrong with my solution and why ?

\(128 = 4^{(2t+1)} – 4^{(t+2)} \)

\(2^7 = 2^{2(2t+1)}- 2^{2(t+2)}\)

\(2^7 = 2^{4t+2} - 2^{2t+4}\) equate bases

\(7 = 4t+2-(2t+4) \)

\(7 = 4t+2-2t-4\)

\(7 = 4t-2t-2\)

\(7 = 2t(2-1)-2\)

\(2t=9 \)

t=9/2

The property you're thinking of is: If x^a = x^b, then a = b (as long as x ≠ 0, x ≠ 1 and x ≠ -1)

There is no similar property that says: If x^a = x^b + x^c, then a = b + c

Consider this example: It's true that 2^4 = 2^3 + 2^3
However, we can't then conclude that 4 = 3 + 3

Cheers,
Brent