How do you solve this simple problem: (-2)^(n-1) = 2^8

[barryj]

How do you solve this..

(-2)^(n-1) = 2^8

The answer is 9 but how do you get it?

 

[Andrew Mason]

How do you solve this..

(-2)^(n-1) = 2^8

The answer is 9 but how do you get it?

You could write the left side as (-1 x 2)^(n-1) = -1^(n-1) x 2^(n-1). Does that help?

AM

 

[barryj]

OK, so continue..what is the proper solution?

 

[Nidum]

Solve by inspection :

The sign must be the same on both sides so (n-1) must be even and (-2)^(n-1) is the same as ... ?

 

[Mark44]

OK, so continue..what is the proper solution?

We're here to help you, but only by guiding you to reach the solution.

 

[Andrew Mason]

You should try to work it out. Since (-1)^8 = 1, you can multiply both sides by 1 this way:

(-2)^(n-1) = (-1)^(8) x 2^8 = (-2)^8

Does that help?

AM

 

[barryj]

This seems to work. However, I wonder if this is considered a general solution. Is there any other way to solve other than knowing that -1^8 = 1? Could I change the problem slightly where this trick (-1^8 = 1) wouldn't work?
I originally thought about using logarithms but you can't take the log of a negative number.

 

[SlowThinker]

I originally thought about using logarithms but you can't take the log of a negative number.

At $\text[ \text{I}\text]$ level this shouldn't be a problem.
$log(-2)=log(2)+\pi i+2 k \pi i$ for $k\in Z$.

 

[Aufbauwerk 2045]

I think perhaps this is a trick question, in the sense that the answer is obvious, if you think about it before just mechanically grinding out the algebra.

The first thing to note is that -2^m = 2^m if m is even.

-2^2 = -2 * -2 = 4
-2^3 = 4 * -2 = -8
-2^4 = -8 * -2 = 16
-2^5 = 16 * -2 = -32
-2^6 = -32 * -2 = 64
-2^7 = 64 * -2 = -128
-2^8 = -128 * -2 = 256 = 2 ^8

This should be clear if you know that multiplying two negative numbers results in a positive number.

Of course you should not need to write the above equations. Really you are just solving for n-1 = 8, therefore n = 9.

In general, if you take the standardized math tests like the SAT or GRE, or even tests in school, sometimes a little thought saves a lot of grind work.

[Edited to correct some typos, which can happen when one is in a hurry to write down the obvious!]

 

[Mark44]

The first thing to note is that -2^m = 2^m if m is even.

No, this isn't true. The left side is a negative number, and the right side is a positive number.

The usual order of operations is that exponents are evaluated before the negation, so -2^2, say is -4, while (-2)^2 is 4.

 

[Mark44]

At $\text[ \text{I}\text]$ level this shouldn't be a problem.

I believe that this is really a B-level thread, not I-level, so I have changed the level to "B".

$log(-2)=log(2)+\pi i+2 k \pi i$ for $k\in Z$.

 

[Aufbauwerk 2045]

No, this isn't true. The left side is a negative number, and the right side is a positive number.

The usual order of operations is that exponents are evaluated before the negation, so -2^2, say is -4, while (-2)^2 is 4.

You mean -2^2 = -(2^2) = -4, or in general -2^(n-1) = -(2^(n-1)).

But he wrote (-2)^(n-1), not -(2^(n-1)).

So applying PEMDAS I interpret this to mean he is applying the exponent n-1 to -2.

 

[Mark44]

You mean -2^2 = -(2^2) = -4, or in general -2^(n-1) = -(2^(n-1)).

But he wrote (-2)^(n-1), not -(2^(n-1)).

PEMDAS.

My comment was relevant to what you wrote; namely

The first thing to note is that -2^m = 2^m if m is even.

and I included your quote in my post (#10).
-2^m is a negative number while 2^m is positive.

 

[Aufbauwerk 2045]

My comment was relevant to what you wrote; namely

and I included your quote in my post (#10).
-2^m is a negative number while 2^m is positive.

I think you are right. I'm used to doing this as I did, since I have thought it's clear from the context. But that does not make it right.

I did some quick "research" to see what Dr. Math says. He would agree with you.

http://mathforum.org/library/drmath/view/55713.html

Thanks for pointing this out. I will correct my answer to the following.

The first thing to note is that (-2)^m = 2^m if m is even.

(-2)^2 = (-2) *( -2) = 4
(-2)^3 = 4 *(-2) = -8
(-2)^4 = (-8) * (-2) = 16
(-2)^5 = 16 * (-2) = -32
(-2)^6 = (-32) * (-2) = 64
(-2)^7 = 64 * (-2) = -128
(-2)^8 = (-128) * (-2) = 256 = 2 ^8

Thanks for the correction.

From Dr. Math:

"When you have numbers only, as in -9^2, it's not at all clear that we
should treat it differently from -x^2. However, some will argue that
it should, because -9 represents a single number, not an operation on
a number. Thus, some will interpret -9^2 as (-9)^2, while others will
read it as -(9^2)."

So I was in the "some will argue" camp. But if there is any possibility of confusion, we should use parentheses. So I am going to accept your correction.

Now I will be on the lookout for this tendency everywhere. :)

 

[Aufbauwerk 2045]

p.s. now it would be interesting to see if, in mathematical papers or textbooks, there is a tendency to distinguish between -x^n and -2^n, in the interpretation sense mentioned by Dr. Math. I suspect the reason I use the convention I do is because I learned it by example, either from books or in the classroom. Very interesting.

 

[Mark44]

The post from Dr. Math that you cited made me think, as well. I've known for many years that -x^2 means the negative of the square of x. I assumed that the same idea applied for constants, as in -2^4. It didn't occur to me that some would distinguish between the negation operation on 2 vs. -2 as a number in its own right.

I looked for some web sites to see if there are varying treatments of, say -2^2. Here's one I found that has a calculator (https://www.symbolab.com/solver/simplify-calculator/simplify -2^{2}), and that treats -2^2 as if it were written -(2^2).

The explanation shown under Steps is horrible.
$2^2 = 4$
$= -4$
Arrrgghh!

 

[UsableThought]

The first thing to note is that (-2)^m = 2^m if m is even.

This should be clear if you know that multiplying two negative numbers results in a positive number.
Of course you should not need to write the above equations. Really you are just solving for n-1 = 8, therefore n = 9.

Reason I noticed this thread is that my current project as an amateur is to review my high school algebra (via a battered copy of Algebra: Structure and Method, by Brown and Dolciani); and in working my way through the first few chapters, I've realized something I probably neither noticed nor cared about 45 years ago - namely how easy it is to get bitten by notation involving negative signs, as has been discussed here. Another interesting example of this (at least to me, 45 years later) is that a negative sign in front of a variable needn't mean the variable itself is negative.

I am curious though about one thing. I agree w/ @Aufbauwerk 2045 about this problem seeming almost deliberately trivial; so @barryj, can I ask where you found it and what the context was?

 

[moriheru]

In reply to the original post: ( I don't know if this has already been mentioned) as the bases are the same on both sides you can equate the exponents. This gives n-1=8 and hence n=9.

 

[Mark44]

In reply to the original post: ( I don't know if this has already been mentioned) as the bases are the same on both sides you can equate the exponents.

The bases aren't the same on both sides. On the left the base is -2, and on the right it's 2.

This gives n-1=8 and hence n=9.