Answer:
Given equation:
[tex]\left(\dfrac{1}{27}\right)^x=3^{(-4x+6)}[/tex]
27 can be written as [tex]3^3[/tex]
Also [tex]\dfrac{1}{a^b}[/tex] can be written as [tex]a^{-b}[/tex]
[tex]\implies \dfrac{1}{27}=\dfrac{1}{3^3}=3^{-3}[/tex]
Therefore, we can rewrite the given equation with base 3:
[tex]\implies (3^{-3})^x=3^{(-4x+6)}[/tex]
To solve, apply the exponent rule [tex](a^b)^c=a^{bc}[/tex]
[tex]\implies 3^{-3 \cdot x}=3^{(-4x+6)}[/tex]
[tex]\implies 3^{(-3x)}=3^{(-4x+6)}[/tex]
[tex]\textsf{If }a^{f(x)}=a^{g(x)}, \textsf{ then } f(x)=g(x)[/tex]
[tex]\implies -3x=-4x+6[/tex]
Add [tex]4x[/tex] to both sides:
[tex]\implies x=6[/tex]
