How do you approach this when it comes to adding or subtracting like bases? My first instinct would to multiply the exponents out and add them, but that can get big and messy. Surely, there's a fast way to approach this?

9^7 - 9^2 = ? and 2^5 + 2^5 + 3^5 + 3^5 + 3^5 = ?

9^7- 9^2= 9^2(9^5-1) [ Take the Common Factor which is 9^2]
= 81(81.81.9-1) [ 9^5= 9^2* 9^2*9 = 81*81*9)


2^5+2^5+3^5+3^5+3^5= 2^5(1+1)+3^5(1+1+1) [ Take Common Factor 2^5, 3^5]
= 2^5.2 + 3^5.3
= 2^6+3^6
= 64+ 729 [(3^2)*3= 9^3= 9^2.9 = 81.9 = 729)


Thanks

When you break it down like that then yes, I agree it can be done. However, the answer choices weren't given from what you had calculated. I don't remember all the choices, but for the first problem,

the answer choices were given in the bases of "9^x"

And that is the same with the 2nd problem where the answer were given in "2^x + 3^x".

Apparently, we have to solve for x. The answer to the second problem was "2^6 + 3^6". Not sure how you get that without a calculator...

Also, I guess the tip I'm wondering or if there is a pattern is how do you apply exponents to addition / subtraction problems?

7^9 - 7^7 = ?

The choices typically given for the answers are 7^x where you have to apparently solve for x. Do you typically just take the common factor ?

When you break it down like that then yes, I agree it can be done. However, the answer choices weren't given from what you had calculated. I don't remember all the choices, but for the first problem,

the answer choices were given in the bases of "9^x"

And that is the same with the 2nd problem where the answer were given in "2^x + 3^x".

Apparently, we have to solve for x. The answer to the second problem was "2^6 + 3^6". Not sure how you get that without a calculator...




Response:-
Ok. When you have to solve for x, identify the BASE. 9^x has the base 9.
Try to convert the given expression into the base equivalent form. For eg, if you have to solve for x given, 9^x= 3^4, then convert 3^4 into Base 9 or 9 into base 3 form.

Here, 3^2x= 3^4 -> 2x=4; x=2


As per the given problem:-

1)
9^7 - 9^2 = 9^2(9^5-1)= 9^x

9^5- 1= 9^x/9^2
9^5-1= 9^(x-2) (Moving the power of 2 in the denominator to the numerator)

x-2= 5 ( to find the value for x, equating the powers of base 9 )

Hence x=7.

2)
2^5 + 2^5 + 3^5 + 3^5 + 3^5 = ?

2^5+2^5= 2.2^5= 2^1.2^5 = 2^6 (this is of the form. a^m * a^n= a^(m+n), since the base "a" is same )

3^5+3^5+3^5= 3^5* 3^3 = 3^8


Can you check if the answer is 2^6 +3^8?

Thanks

the answer was 2^6 + 3^6

In your problem, you said

1)
9^7 - 9^2 = 9^2(9^5-1)= 9^x

9^5- 1= 9^x/9^2
9^5-1= 9^(x-2) (Moving the power of 2 in the denominator to the numerator)

x-2= 5 ( to find the value for x, equating the powers of base 9 )

Hence x=7.

What happen to the "-1" when you cancel out the bases of 9?

x-2 = 5 but what happen to the "-1" ?

the main idea is in sudhan's post, up a couple of posts from this one. you HAVE to get used to the idea that multiplying 2^n by another '2' gives 2^(n+1), because the extra '2' can be rewritten as 2^1 (whereupon the standard exponent rules apply). this just might be the most commonly applied exponent trick on the exam.
they do it twice in this problem alone:
2 x 2^5 = 2^6
3 x 3^5 = 3^6
know it!

sudhan:



the second line here is incorrect: 3^5 + 3^5 + 3^5 is (3)(3^5), which is 3^6. the term '3^5' only appears 3 times, not 3^3 times.
two weird things about this:
1) you didn't make the same mistake in the first line;
2) your first post on this thread does this exact part of the problem correctly (!)

Thanks Ron for pointing my mistake.

Thanks for the help. I understand what Sudhan is trying to illustrate. Could you guys try to explain the other part of the question ? What happen to the "-1"?

1)
9^7 - 9^2 = 9^2(9^5-1)= 9^x

9^5- 1= 9^x/9^2
9^5-1= 9^(x-2) (Moving the power of 2 in the denominator to the numerator)

x-2= 5 ( to find the value for x, equating the powers of base 9 )

Hence x=7.

Other than that part, I'm crystal clear. :)

you can't get rid of that 1; it is going to stay there, unless you're forgetting some mathematical operation that you're supposed to perform.
in fact, if a and b are integers, it is actually for 9^a - 9^b to be another power of 9 - or, in fact, for this to happen for any other integer other than 9.
(note that if you could combine different powers in general, then there would be no such thing as a polynomial!)

in any case - you should go back to the original source (gmatprep), find the original problem, and post it here; we'll see what the difference is. there is absolutely no way you are going to turn 9^7 - 9^2 into a single power of 9.