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Sagot :
Let's solve ourselves instead of believing anyone
[tex]\\ \rm\Rrightarrow (x^2)^7=x^4x^8[/tex]
- a^m+a^n=a^m+n
[tex]\\ \rm\Rrightarrow x^{14}=x^{4+8}[/tex]
[tex]\\ \rm\Rrightarrow x^{14}=x^{12}[/tex]
[tex]\\ \rm\Rrightarrow x^{14}-x^{12}=0[/tex]
[tex]\\ \rm\Rrightarrow x^{12}(x^2-1)=0[/tex]
[tex]\\ \rm\Rrightarrow x^2-1=0[/tex]
[tex]\\ \rm\Rrightarrow x^2=1[/tex]
[tex]\\ \rm\Rrightarrow x=\pm 1[/tex]
- 0 is also a solution
Answer:
Joe is correct
Step-by-step explanation:
Given equation:
[tex](x^2)^?=x^4 \cdot x^8[/tex]
The exponent outside the bracket is a question mark and the students are trying to determine the value of the question mark.
For ease of answering, let y be the unknown number (question mark):
[tex]\implies (x^2)^y=x^4 \cdot x^8[/tex]
First, simplify the equation by applying exponent rules to either side of the equation:
[tex]\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}\quad\textsf{to the left side}:[/tex]
[tex]\implies (x^2)^y=x^4 \cdot x^8[/tex]
[tex]\implies x^{2y}=x^4 \cdot x^8[/tex]
[tex]\textsf{Apply exponent rule} \quad a^b \cdot a^c=a^{b+c} \quad \textsf{to the right side}:[/tex]
[tex]\implies x^{2y}=x^4 \cdot x^8[/tex]
[tex]\implies x^{2y}=x^{4+8}[/tex]
[tex]\implies x^{2y}=x^{12}[/tex]
Now, apply the exponent rule:
[tex]x^{f(x)}=x^{g(x)} \implies f(x)=g(x)[/tex]
Therefore,
[tex]x^{2y}=x^{12}\implies 2y=12[/tex]
Finally, solve for y:
[tex]\implies 2y=12[/tex]
[tex]\implies 2y \div 2 = 12 \div 2[/tex]
[tex]\implies y=6[/tex]
Therefore, Joe is correct as the unknown number is 6.
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