Exponentiation mistake

Consider $(-a)^b$ . If $b$ is odd, then $(-a)^b$ will always be negative (Make sure you see why). If $b$ is even, then $(-a)^b$ will always be positive, since the negative signs will cancel out.

Now consider $a^b$ . No matter what, if both $a$ and $b$ are positive (as stated in the problem), $a^b$ will be positive. This is because a positive number raised to any real number is positive.

So, we know that if $b$ is odd, $(-a)^b \neq a^b$ , since a number cannot be both negative and positive. However, we can notice that if $b$ is positive, $(-a)^b$ will always be equal to $a^b$ , since the exponents stay the same and the negative signs cancel out.

So, we have that $a$ can be anything (1-10, 10 choices) and $b$ must be even (5 even numbers from 1-10), for a total of $5\times 10 = \boxed{50}$ ordered pairs of integers $(a,b)$ that satisfy $(-a)^b = a^b \ \blacksquare$

$(-a)^b=a^b$ when $b$ is an even number. So possible values of $b$ are $2,4,6,8,10$ and $a$ can be any value between $1$ to $10$ . Answer is $10*5=\boxed{50}$

We know that in order to make a negative integer positive, we have to square the integer with even numbers only. This means that the possible b-values are $b = 2$ , $4$ , $6$ , $8$ , and $10$ .

At the same time, since the negative has been 'made positive' after squaring with even numbers, we are left with $a^b = a^b$ . The equation holds for all possible values of a, in which this case means $a = 1, 2, ..., 10$ . Therefore the number of ordered pairs of $(a,b)$ is $10 \times 5 = \boxed{50}$

$(-a)^b=(-1)^b \times a^b$ .

$\Rightarrow$ $(-1)^b \times a^b = a^b$ .

As $a^b \neq 0$ , The equation satisfies when $(-1)^b=1$

which is the case when $b$ is even.So,there are 10 possibilities for $a$ and 5 for $b$ .

$\Rightarrow$ Number of ordered pairs = $10 \times 5 = \boxed{50}$

It's clear that any even exponent $b$ will result in $(-a)^b=a^b$ , while any odd exponent will result in $(-a)^b = -a^b$ . There are 5 even integers on $[1,10]$ , so $5 \times 10 = 50$ .

If $b$ is odd, then $(-a)^b=-a^b\neq a^b$ so $b$ cannot be odd.

Thus, $b$ is even, and $(-a)^b=a^b$ is true for any $a$ .

$b$ can be any of the 5 even integers in the range, and $a$ can be any of the 10 integers in the range, and so the answer is $5\cdot 10=\boxed{50}$ .

Note that for all even $b$ , we have $(-a)^b=a^b$ because $(-a)^b=(-1)^b\cdot a^b=a^b$ when $b$ is even. Therefore we can pick $2,4,6,8,10$ , or $5$ different numbers for $b$ , and $10$ different numbers for $a$ giving a grand total of $5\times 10=\boxed{50}$ pairs.

To find the number of ordered pairs $(a,b)$ ,we must multiply the number of possibilities for $a$ and $b$ .

I think that find the number of possibilities for $b$ first is easier.

We can find that $(-a)^b=a^b$ sometimes,but only when $b$ is an even number .

Why?

It is because that a negative number times a negative number equals to a positive number.Example,assume that integers $-x$ multiplies $-y$ ,the answer will be

$-x\times-y=-(-(xy))=xy$

The answer is same to $x\times y=xy$

Same.If $b$ is even,then $(-a)^b$ equals to $a^b$ .

(If $b=2$ ,then $(-a)^2=-a\times-a=-(-(a^2))=a^2)$

The problem says that integers $(a,b)$ each are between $1$ and $10$ inclusive.

We know that $b$ can be any even number from $1$ to $10$ ,so there are $5$ possibilities for $b$ . $(2,4,6,8,10)$ .

Okay,let's do the next step.

How many possibilities are there for $a$ ?It is just simple, $a$ can be any number from $1$ to $10$ .Therefore,there are $10$ possibilities for $a$ .

Now,we know the number of possibilities for $a$ and $b$ .Let's calculate our answer.

From above,there are 10 possibilities for $a$ and 5 possibilities for $b$ .

Number of ordered pairs= $10\times5=\boxed{50}$ pairs

This is fairly straightforward to solve with simple analysis.

For one, you know that if the a value and the b value are the same on both sides, which they are, the absolute value of the number will be the same on both sides no matter what. This sounds fairly obvious, but it has an important implication: any value for a is perfectly acceptable, especially since we can only use values 1-10 inclusive, meaning no negative values for a.

Now we simply remember our properties of exponents: any even exponent will create a positive number, regardless of if the even exponent is on a positive or a negative number. This is due to the fact that a negative times a negative is positive. We know between 1 and 10, there are five even numbers (2, 4, 6, 8, 10), meaning only five possible values for b.

So, combine these two facts: all ten values work for a, and all five values work for b. Any values of a can be paired with any working value of b. Therefore, in order to get the number of possible combinations, we multiply our number of possible a values and our number of possible b values: 5 * 10 = 50, our answer.

You can solve, with simple thinking. First, we know that the power of negative numbers will be same with the power of positive numbers if only if the power is even numbers. So, we can choose $a$ from 1 to 10 (inclusive). Then, the power indicated $b$ , you must choose the even numbers from 1 to 10 (inclusive), it 's five even numbers. Finally, there are 10 numbers of $a$ and 5 of even numbers $b$ , we can choose randomly. $(a, b)$ . Answer is multiplication rule , $10 * 5$ = $\boxed{50}$ .

The assertion Murray makes is only true if $b$ is even (if $b$ is odd, $(-a)^{b} = -a^{b}$ ). $b$ is independent of $a$ , and $50\%$ of $b$ 's potential values are even. Therefore, $\boxed{50}/100$ possibilities satisfy the statement.

$(-a)^{b} = a^{b}$ is true only when $b$ is an even number. Hence, there are $5$ such possible values of $b$ .

However, $a$ has no such restriction. $a$ can take any of the $10$ possible values,

So, number of ordered pairs $a$ , $b$ is $10 \times 5 = 50$

When (-a)^b is not equal to a^b ???

If b is odd. Because then (-a)^b is a negative number but a^b is always a positive number.

Now, how many ordered pair are their ???

For every 'a'(from 1 to 10) we get 5 even numbers to satisfy the above equation. So answer is 10*5=50. :)

A=1,2,3,4,5,6,7,8,9,10 B=2,4,6,8,10 5times10=50

(-a)^b=(-1)^b X (a)^b so b has to be has to be positive. b is even . so b can take 5 values(2,4,6,8,10). a can take all values within domain. so total pairs 5X10=50

b is the governing number if b is even its immaterial, what is value of a the equation would be satisfied. (1,2) (2,2) .................... (10,2) (1,4) (2,4)......................(10,4) . . . . . . (10,2) (10,4)............................(10,10)

For this problem, all even values for b (the exponent) are correct. e.g. let a = 3, let b = 2 (-3)^{2} = (-3) x (-3) = 9 and 3^{2} = 3 x 3 = 9, thus under these circumstances, (-3)^{2} = 3^{2}, and (- a )^{ b } = a ^{ b }, using reflexive property The same result occurs for all values of a 1 through 10, and for the same value of b .

There are then a total of five even values for b , all of which work the same way as the example above, just with more numbers, and each has ten different possibilities for a . So, 5 potential b values times 10 a values for each of those b values gives you 50 total ordered pairs.

first write it

$$ (-1)^b a^b = a^b \Rightarrow (-1)^b = 1 \Rightarrow 1 \leq b = 2n \in \Bbb N \leq 10 \Rightarrow b = 2 \space , 4 \space , 6 \space 8 \space , 10 $$

and $$ 1 \leq a \in \Bbb N \leq 10 $$ $$ | {a : 1 \leq a \in \Bbb N \leq 10} | = 10 $$ $$ \Rightarrow \mathrm{the \space number \space of \space pairs} = 5 \times 10 = 50 $$

We can write $(-a)^b$ as $(-1)^b*a^b$

We are required to find solutions of $(-a)^b = a^b$

$=> (-1)^b*a^b = a^b$

$=> (-1)^b = 1$ [Since $a \ne 0$ ]

$=> b$ is an even number

Since $b$ is between 1 and 10 inclusive, $b$ can be any of the 5 even numbers.

$a$ can be any number between 1 and 10 inclusive.

For each choice of $a$ , we have 5 choices of $b$ . As there are 10 choices of $a$ : Number of ordered pairs = 10*5 = 50

This must be true when "b" is an even number between 1 to 10 i.e 2,4,6,8,10 & for this a could be any number between 1 to 10. so total number of pairs 10*5=50

all even numbered exponential terms are even functions f(a)=f(-a)

We can write $(-a)^b=(-1)^b\times a^b$ . For this to be equal to $a^b$ , we have $(-1)^b\times a^b=a^b\Leftrightarrow (-1)^b=1$ . Clearly, this is only true when $b$ is even. So the only restriction is that $b$ is even, and then $a$ can take any value. So our pairs are of the form $(a,\{2,4,6,8,10\})$ and there are exactly $10$ choices for $a$ and $5$ choices for $b$ , for a grand total of $10\times 5=\boxed{50}$ ordered pairs.

Para que a igualdade sea verdadeira o expoente tem de ser par. b = 2,4,6,8 0u 10 Para o expoente 2 há 10 soluções, ao todo há 5 x 10 = 50 (1,2) (2,2) (3,2)...

Given relation: $(-a)^b= a^b$

Now for the value of $(-a)^b$ to be positive, $b$ must be positive. Therefore, $b$ can have only the following values: $2, 4, 6, 8, 10$ .

But the base can be anything between 1 to 10 since they'll all have positive powers to make their values positive. So, possible values of a: $1, 2, 3, 4, 5, 6, 7 , 8, 9, 10$ .

The total number of possible relations is hence given by : $5 \times 10 = \boxed{50}$ .

Let's begin with a specific example.

If

$b = 2$ , the equation holds true, because $(-a)^2 = a^2$ . The equation thus holds true for any a between 1 and 10 inclusive. This gives us 10 ordered pairs:

$[1,2], [2,2], [3,2], ... , [10,2]$ .

Realizing that any even integer between one and ten will actually still result in

$(-a)^b = a^b$

and knowing that we have 4, 6, 8, 10 with ten ordered pairs for each number, we have

$10 \times 5 = \boxed {50}$