Let's address each part of the problem step-by-step.
Part (a): Value of [tex]\( x^0 \)[/tex]
Recall that any non-zero number raised to the power of zero is equal to 1. This is a fundamental property of exponents.
So, for any [tex]\( x \neq 0 \)[/tex]:
[tex]\[ x^0 = 1 \][/tex]
Thus, the value of [tex]\( x^0 \)[/tex] is [tex]\( 1 \)[/tex].
Part (b): Proof that [tex]\( (2x + 3y)^{2m} \times (2x + 3y)^{2n} = 1 \)[/tex] given [tex]\( m + n = 0 \)[/tex]
We are given:
[tex]\[ m + n = 0 \][/tex]
We need to prove:
[tex]\[ (2x + 3y)^{2m} \times (2x + 3y)^{2n} = 1 \][/tex]
First, let's look at the properties of exponents. One useful property is:
[tex]\[ a^b \times a^c = a^{b+c} \][/tex]
Applying this property to our terms:
[tex]\[ (2x + 3y)^{2m} \times (2x + 3y)^{2n} = (2x + 3y)^{2m + 2n} \][/tex]
We know that:
[tex]\[ 2m + 2n = 2(m + n) \][/tex]
Since [tex]\( m + n = 0 \)[/tex], we substitute this in:
[tex]\[ 2(m + n) = 2(0) = 0 \][/tex]
Thus, the exponent becomes:
[tex]\[ (2x + 3y)^{2m + 2n} = (2x + 3y)^0 \][/tex]
And since any number (except zero) raised to the power of zero is 1, we have:
[tex]\[ (2x + 3y)^0 = 1 \][/tex]
Therefore:
[tex]\[ (2x + 3y)^{2m} \times (2x + 3y)^{2n} = 1 \][/tex]
Conclusion:
(a) The value of [tex]\( x^0 \)[/tex] is [tex]\( 1 \)[/tex].
(b) If [tex]\( m + n = 0 \)[/tex], then:
[tex]\[ (2x + 3y)^{2m} \times (2x + 3y)^{2n} = 1 \][/tex]
We have shown this using the properties of exponents and the given condition.
