Breaking Down a GMATPrep Exponential Equations Problem
This week, were going to tackle a harder GMATPrep problem solving question from the topic of Exponential Equations (a subset of Algebra).
Lets start with the problem. Set your timer for 2 minutes. and GO!
*The temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees Fahrenheit. If the temperature F of the coffee t minutes after it was poured can be determined by the formula [pmath]F=120(2^{-at})+60[/pmath], where F is in degrees Fahrenheit and a is a constant, then the temperature of the coffee 30 minutes after it was poured was how many degrees Fahrenheit?(A) 65
(B) 75
(C) 80
(D) 85
(E) 90
The problem gives us a formula; thats probably the first thing that most people will write down. The formula contains three variables: F (for temperature), t (for time) and a (an unknown constant). The problem also gives one particular scenario for a temperature, F, at a certain time, t.
Hey, those are two of the three variables! Thats great we can just plug them in and solve for a! Then well know the constant. And then we can plug that into the main formula to solve another time for the answer.
If you were writing a math textbook, or trying to solve using official math processes, then this is exactly what youd want to do. But we dont care about that. We just care about getting to the right answer in the easiest possible way. So lets talk about the official math way first and then discuss how we could use that understanding to discover an easier way.
Also, whenever you have a problem that might include conversions (hours to minutes, for example, or Fahrenheit to Celsius), double-check everything to see whether you do have any. In this case, time is always in minutes and temperature is always in Fahrenheit, so we dont need to worry about any conversions on this problem.
So, we now have this:
[pmath]120=120(2^{-a10})+60[/pmath]
[pmath]120-60=120(2^{-a10})[/pmath]
[pmath]60/120=(2^{-a10})[/pmath]
Note that we have a negative in the exponent on the right-hand side of the equation. How do we get rid of negatives in exponents? Take the reciprocal of the base term, so
[pmath](2^{-a10})[/pmath] becomes [pmath]1/2^a10[/pmath]
Now we have:
[pmath]1/2^1=1/2^a10[/pmath]
The bases (1/2) are the same, so we can drop the bases and set the exponents equal to each other:
[pmath]1=10a[/pmath]
[pmath]a=1/10[/pmath]
Now we know the value for the constant a: 1/10 or 0.1. Our standard formula becomes:
[pmath]F=120(2^(-0.1t))+60[/pmath]
The problem asks us to find the temperature, F, when the time, t, is 30 minutes:
[pmath]F=120(2^({-0.1}*30))+60[/pmath]
[pmath]F=120(2^{-3})+60[/pmath]
[pmath]F=120(1/8)+60[/pmath]
[pmath]F=15+60[/pmath]
[pmath]F=75[/pmath]
And were done! The correct answer is B.
Which part of that did you find the most annoying? For me, and for most people, it was having to solve for a. It wasnt absolutely terrible, but Id like to avoid having to do that math if I can.
When were first reading and writing down the problem, we need to realize several things:
(1) theyve given us the standard formula with three variables
(2) theyve given us an instance in which we can use two values for two of the variables (a temperature of 120 and a time of 10 minutes) in order to solve for the third (the constant)
(3) they are not actually asking us to solve for that third variable, the constant, and
(4) they are asking us to solve for another instance, or usage, of the original formula (the temperature at a time of 30 minutes)
Remember this?
[pmath]120=120(2^{-a10})+60[/pmath]
This was our starting point for the math we did above. Before you start using that formula to solve for a, also write out the one you definitely need to solve:
[pmath]F=120(2^{-a30})+60[/pmath]
You want to solve for F. Whats the painful and unnecessary spot? That constant, a. Im not asked to find a, and yet it looks like I have to solve for a in order to get the answer. Examine the two equations. Theyre very similar. and the other numbers in the problem are not that complex maybe we can find an easier way to get what we need here. Lets isolate the annoying term, [pmath](2^{-a10})[/pmath], in our first equation, just as we did earlier:
[pmath]120-60=120(2^{-a10})[/pmath]
[pmath]60=120(2^{-a10})[/pmath]
[pmath]1/2=(2^{-a10})[/pmath]
And, now, lets not solve for a, okay? We have a specific value for that nasty term that includes the unknown a. I actually need a value for a very similar (and equally nasty) term.
How do I go from [pmath](2^{-a10})[/pmath] to [pmath](2^{-a30})[/pmath]?
I have an exponent of 10 and I want an exponent of 30. If I raise the exponent to another exponent, 3, then that 10 turns into 30! For example, if I had [pmath]x^10[/pmath], I could raise that to the power of three: [pmath](x^10)^3[/pmath]. That would give me [pmath]x^30[/pmath]. Back to our problem. Whatever I do to one side of the equation, I have to do to the other, right?
[pmath](1/2)^3=[(2^{-a10})]^3[/pmath]
[pmath]1/8=[(2^{-a30})][/pmath]
Who cares what a is? We have a value to plug in for that nasty term; thats all we care about.
[pmath]F=120(2^{-a30})+60[/pmath]
[pmath]F=120(1/8)+60[/pmath]
[pmath]F=15+60[/pmath]
[pmath]F=75[/pmath]
The last three lines are identical to our first solution method we just got to skip the painful part in the middle.
Key Takeaways for Solving Hard Equation Problems:
- You may have done the long algebra the first time and gotten the problem right confidently in 2 minutes (or faster!) but that doesnt mean that you still cant learn a more efficient way to do the problem. If something feels especially annoying, long, or problematic, there probably is a more efficient way. While studying, whenever I finish a problem, I always ask myself, What was the most annoying part of that solution? and then I try to find a way to avoid having to do that! (Note: if youre taking the test and you dont know the easier way, dont try to figure it out then. Just do it however you can.)
- When you have multiple versions of the same formula, and it looks like you have to solve for an intermediate variable in order to get the final, desired variable look again, especially if it would be annoying to find that intermediate variable.
- If you have a value for one annoying term and you want to find a value for a similar term, ask yourself what math you have to do to go from the first form of the annoying term to the second. Then apply that same conversion to the value.
* GMATPrep questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.