k = 3 + 6x
note that 27 = 3³ and 9 = 3²
27 × [tex]9^{3x}[/tex] = 3³ × (3²)^3x = 3³ × [tex]3^{6x}[/tex] = [tex]3^{3 + 6x}[/tex]
hence k = 3 + 6x
Answer:
[tex]k=3+6x[/tex].
Step-by-step explanation:
We haven been given an expression [tex]27\times 9^{3x}[/tex]. We are asked to write our given expression is form [tex]3^k[/tex] and find out value of k.
We can rewrite 27 as [tex]3^3[/tex] and 9 as [tex]3^2[/tex]. Upon substituting these values in our given expression, we will get:
[tex]3^3\times (3^2)^{3x}[/tex]
Using exponent property [tex](a^b)^c=a^{b\cdot c}[/tex], we will get:
[tex]3^3\times 3^{2\cdot 3x}[/tex]
[tex]3^3\times 3^{6x}[/tex]
Using exponent property [tex]a^b\times a^c=a^{b+c}[/tex], we will get:
[tex]3^{3+6x}[/tex]
We have written our given expression in form [tex]3^k[/tex]:
[tex]3^{k}=3^{3+6x}[/tex]
Therefore, the value of k is [tex]3+6x[/tex].
