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exponents problem

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Source: — Problem Solving |

by scoobydooby » Tue Jun 16, 2009 9:54 am
2^(x) - 2^(x-2) = 3(2^13)

=>2^(x-2){2^2-1}=3(2^13) (taking 2^(x-2) common on the LHS)
=>2^(x-2)*3=3(2^13)
=> x-2=13 (comparing the powers on both sides)
=>x=15
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Re: exponents problem

by Stuart@KaplanGMAT » Tue Jun 16, 2009 9:55 am
LFalken wrote:if 2^(x) - 2^(x-2) = 3(2^13)

what is the value of X? could you please show your work?

answer is equal to 15, Thanks!
If we want to add or subtract powers, we must have the same base AND the same exponent. Once we get our terms in similar form, we simply add the coefficient.

For example, 5(x^3) + 3(x^3) = 8(x^3)

To solve the problem you posted, we need to start by equalizing the exponent of the two terms on the left side of the equal sign. Let's ignore the right side for now (it's not going anywhere!).

2^x - 2^(x-2)

In general, we'll lower the bigger exponent to match the smaller one. Here, our bigger exponent is "x", so we want to convert

2^x

to (something)^(x-2).

We do so using our exponent multiplication rule:

a^(y+z) = (a^y)(a^z)

Accordingly:

2^x = 2^(x-2) * 2^2

since (x-2) + 2 = x

Substituting in for 2^x, we now have:

2^(x-2) * 2^2 - 2^(x-2)

4*2^(x-2) - 2^(x-2)

Note that this is essentially the same as 4y - y, in which y=2^(x-2), so we simply subtract the coefficients:

4*2^(x-2) - 2^(x-2) = 3*2^(x-2)

Almost done! Now we'll reattach to our lost lost right side of the equation:

3(2^(x-2)) = 3(2^13)

and we should be able to see that the coefficient is the same on both sides ("3"), the base is the same on both sides ("2"), which means that the exponent also must be the same on both sides:

x-2 = 13

x=15!

Of course, it takes a LOT longer to explain this than it should to do the actual math - once you understand the concept, you can attack these questions very quickly.

* * *

As a complete aside, these questions are also great for backsolving - you can just plug in the answers for x so you're working with hard values instead of variables and it makes the whole thing easier to solve.
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by LFalken » Tue Jun 16, 2009 10:43 am
thanks to both of you! the detailed explanation was particularly helpful...
LFAL
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by LFalken » Tue Jun 16, 2009 10:47 am
thanks to both of you! the detailed explanation was particularly helpful...
LFAL
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