Answer:
Image for the above Question is Missing hence Attached below.
[tex]y^{2}y^{4}[/tex] ⇒ [tex]y^{6}[/tex]
[tex](2y)^{2}[/tex] ⇒ [tex]4y^{2}[/tex]
[tex](\dfrac{y}{4})^{2}[/tex] ⇒[tex]\dfrac{y^{2}}{16}[/tex]
[tex]2((y)^{3})^{3}[/tex] ⇒ [tex]2y^{9}[/tex]
Step-by-step explanation:
We have Law of Indices as follow
1. [tex]x^{a}x^{b}=x^{a+b}[/tex]
2. [tex](xy)^{a}=x^{a}y^{a}[/tex]
3. [tex](\dfrac{x}{y})^{a}=\dfrac{x^{a}}{y^{a}}[/tex]
4. [tex]((x)^{a})^{b}=x^{a\times b}[/tex]
Using above identities we get
For First
[tex]y^{2}y^{4}=y^{2+4}=y^{4}[/tex]
Therefore [tex]y^{2}y^{4}[/tex] ⇒ [tex]y^{6}[/tex]
For Second
[tex](2y)^{2}=2^{2}y^{2}=4y^{2}[/tex]
Therefore [tex](2y)^{2}[/tex] ⇒ [tex]4y^{2}[/tex]
For Third
[tex](\dfrac{y}{4})^{2}=\dfrac{y^{2}}{4^{2}}=\dfrac{y^{2}}{16}[/tex]
Therefore [tex](\dfrac{y}{4})^{2}[/tex] ⇒[tex]\dfrac{y^{2}}{16}[/tex]
For Fourth
[tex]2((y)^{3})^{3}=2y^{3\times 3}=2y^{9}[/tex]
Therefore [tex]2((y)^{3})^{3}[/tex] ⇒ [tex]2y^{9}[/tex]
