Solve the following elementary exponential equation.

[tex]\left(\frac{1}{3}\right)^{4x+4} = 27^x[/tex]

[tex]x = [/tex]



Answer :

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]