To solve the equation [tex]\(\left(\frac{1}{3}\right)^{4x+4}=27^x\)[/tex], we shall follow a step-by-step approach.
1. Rewrite both sides with the same base if possible.
The term [tex]\(\frac{1}{3}\)[/tex] can be rewritten as [tex]\(3^{-1}\)[/tex].
So, [tex]\(\left(\frac{1}{3}\right)^{4x+4}\)[/tex] can be written as [tex]\(\left(3^{-1}\right)^{4x+4}\)[/tex].
This simplifies to:
[tex]\[
3^{-(4x+4)}.
\][/tex]
The term [tex]\(27\)[/tex] can be written as [tex]\(3^3\)[/tex].
So, [tex]\(27^x\)[/tex] can be written as [tex]\(\left(3^3\right)^x\)[/tex].
This simplifies to:
[tex]\[
3^{3x}.
\][/tex]
2. Set the exponents equal to each other since the bases are the same.
Now we have the equation:
[tex]\[
3^{-(4x+4)} = 3^{3x}.
\][/tex]
Since the bases are the same, we can set the exponents equal to each other:
[tex]\[
-(4x + 4) = 3x.
\][/tex]
3. Solve for [tex]\(x\)[/tex].
First, distribute the negative sign on the left side:
[tex]\[
-4x - 4 = 3x.
\][/tex]
Next, combine like terms by adding [tex]\(4x\)[/tex] to both sides:
[tex]\[
-4 = 7x.
\][/tex]
Now, solve for [tex]\(x\)[/tex] by dividing both sides by 7:
[tex]\[
x = -\frac{4}{7}.
\][/tex]
So, the solution to the equation [tex]\(\left(\frac{1}{3}\right)^{4x+4} = 27^x\)[/tex] is:
[tex]\[
x = -\frac{4}{7}.
\][/tex]
