Answer:
[tex]x=-10[/tex]
Step-by-step explanation:
This deals with negative exponents.
[tex]x^{-2}=\frac{1}{x^2}[/tex]
27 and 9 are both powers of 3, so we need to find a way to make both of these into a base 3.
9 is easy, thats just:
[tex]9=3\times3=3^2[/tex]
1/27 is a little more difficult. Start with 27:
[tex]27=9\times3=3\times3\times3=3^3[/tex]
and because its a 1/27, that exponent has to be negative.
[tex]3^3=27\\3^-^3=\frac{1}{27}[/tex]
Now we can get both of them to base 3, so do that in the given equation:
[tex]3^{(-3)\times(4-x)}=3^{(2)\times(2x-1)}[/tex]
Same as last time, the bases can be canceled out as the exponents have to be equal for this equation to be true.
[tex]3^{(-3)\times(4-x)}=3^{(2)\times(2x-1)}\\(-3)\times(4-x)=(2)\times(2x-1)[/tex]
Finally, solve for x:
[tex]-3(4-x)=2(2x-1)\\-12+3x=4x-2\\-10+3x=4x\\x=-10[/tex]
