The temperature inside a certain industrial machine

[chris.innes1]

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

Why can't you set both sides equal to each other with a base of 2, drop the base and solve?

128 = 2 to the 7th

[chris.innes1]

The post above should read 4 raised to 2t +1 minus 4 raised to t +2

[RonPurewal]

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

Why can't you set both sides equal to each other with a base of 2, drop the base and solve?

128 = 2 to the 7th

I fixed the formula for you.

Even if you switch everything to have 2 as the base number, you still have a sum of 2 powers on one side of the equation. In that case, you can't drop the bases, because there's no way to combine 2^stuff + 2^(other stuff) into a single exponential expression.

As a perfect analogy, consider the Pythagorean theorem (a^2 + b^2 = c^2).
Your question here is like asking why we can't take the square root (or "drop the squares") from that formula. If there were one square on each side (and you knew the signs were positive), you could do that -- but not with the sum.

[sirous.afshar88]

If we factor down to a 2, we have an equation that looks like so:

2^7 = 2^4t+2 - 2^2t-4

I thought when we have an equation where all the bases are the same, we can drop the base and work with the powers. However, is this ONLY the case when there isn't any artithmatic needed on either side of the equation (ie + -) as far as the bases are involved ?

Appreciate the clarification.

[chaitanyamisal]

How do you solve such a question in 120 seconds?

[georgepa]

If you can't figure out the algebra - plug in the numbers - and you should be able to do this in under 2 minutes. You only have 5 choices and in the worst case you only have to try 2.

It can help to notice certain things - although even without noticing these you can solve the problem.

1. 128 = 2^(7) = 4^(3) x 2.


2. The right exponent is smaller than the left exponent, i.e 2t + 1 > t+2 for all t > 1.


3. Since the bases are the same , h(t) increases as t increases for any t > 1.
   
   
   
   • e.g. t=2 => h(2) = 4^(5) - 4^(4) = 4^(4)[4^1 - 1]
   
   
   • e.g. t=3 => h(3) = 4^(7) - 4^(5) = 4^(4)[4^3 - 4^2] > h(2)
   
   
   • e.g. t=4 => h(4) = 4^(9) - 4^(6) = 4^(4)[4^5 - 4^3] > h(3)

If you were choosing smartly - you could choose c first.
Since h(5/2) > 128 discard c,d, and e. You could next try b and since h(2) > 128 - you would simply choose a as the answer.

a) t = 3/2

Code: Select All Code

h(3/2) = 4^(4) - 4^(7/2)
       = 2^(8) - 2^(7)
       = 2^(7) [ 2-1 ]
       = 2^(7) = Right choice


b) t = 2

Code: Select All Code

h(2) = 4^(5) - 4^(4)
       = 4^(4) [ 4 -1 ]
       = 4^(4) (3) > 4^(3) x 2


c) t = 5/2

Code: Select All Code

h(5/2) = 4^(2t+1) - 4^(t+2)
       = 4^(5 +1) - 4^(9/2)
       = 2^(12) - 2^(9)
       = 2^(9) [ 2^(3) -1 ]  > 2^(7)


d) t = 3

Code: Select All Code

h(3) = 4^(7) - 4^(5)
       = 4^(5) [ 4^(2) -1 ] > 4^(3) x 2


e) t = 7/2

Code: Select All Code

h(7/2) = 4^(8) - 4^(11/2)
      = 2^(16) - 2^(11)
      = 2^(11) [ 2^(5) -1 ] > 2^(7)


[RonPurewal]

Yep. Plug in numbers.

By the way, "120 seconds" is an average time. If you think of 120 seconds as a LIMIT, you will score lower on this test as a result.
It's not a limit. It's an average.

The best answer, though, is this: If your time management is sufficiently developed, you shouldn't even have to think about numbers of minutes or seconds at all.
Fully developed time management is just 3 things:
1/ Be totally, completely, brutally honst with yourself about when you are STUCK. (If you do not know exactly what you are doing and exactly why you're doing it, then you are stuck.)
2/ When you're stuck, pause, and try to figure out exactly what you are doing and exactly why you're doing it.
3/ If you can't, QUIT.
... If you can think of another approach, IMMEDIATELY try it.
... If you can't, guess and move on.

That's it.

If you have the self-discipline to do these things (especially #1), then you'll never need to think about minutes or seconds again (and you'll find that the "time pressure" on this exam is mostly an illusion).

[MichelleU634]

I have the same question as the previous guy that wasn't answered.

why is this wrong?

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


I thought if you got the same base you could just work with the exponents?

[RonPurewal]

you can only do what you're talking about if 4^this = 4^that. if there's more than one power of 4, you have to try (with algebra, or exponent rules, or whatever) to reduce to one power of 4 on each side.

in other words, you could only write "2t+1-(t+2) = 3+3" if you actually had 4^(2t + 1 - (t + 2)) = 4^(3 + 3)

[RonPurewal]

also, it might just be the algebra (i.e., the presence of a variable) that's making this seem weird.

if that's the case, then it's not hard to construct counterexamples with numbers.
e.g., it's true that 2^5 – 2^4 = 2^3 + 2^3 (both sides are 16), but this certainly does not mean that 5 – 4 = 3 + 3.

[AdhiK295]

In order to solve this we can assume 4^t or 2^2t as variable x:

128 = 4x^2 -16 x
0 = x^2 - 4x - 32
0 = (x-8) (x+4)
2^2t = 8
t = 3/2 (a)

[tim]

That would totally work if you're comfortable substituting variables.

[RonPurewal]

In order to solve this we can assume 4^t or 2^2t as variable x:

128 = 4x^2 -16 x
0 = x^2 - 4x - 32
0 = (x-8) (x+4)
2^2t = 8
t = 3/2 (a)

do you have a question?

if so, please clarify. thanks.