Question

1/3^(3x-2)=81^(4x+1)

Answer 1

$( \frac{1}{3} )^{3x-2}=81^{4x+1}\\ 3^{-(3x-2)}=(3^4)^{4x+1}\\ 3^{-3x+2}=3^{16x+4} \\ -3x+2=16x+4 \\ -3x-16x=4-2 \\ -19x=2 \\ 19x=-2 \\ x=- \frac{2}{19}$

Answer 2

$(\frac{1}{3})^{3x-2}=81^{4x+1} \\ 3^{-3x+2}=3^{16x+4} \\ -3x+2=16x+4 \\ -19x=2 \\ x=-\frac{2}{19}$

Answer 3

To solve the equation $\frac{1}{3^{3x-2}} = 81^{4x+1}$, we need to rewrite $81$ as a power of $3$ because $81 = 3^4$. Substituting this, the equation becomes:

$\frac{1}{3^{3x-2}} = (3^4)^{4x+1}$

Now, apply the rule that says when raising a power to another power, you multiply the exponents:

$\frac{1}{3^{3x-2}} = 3^{(4)(4x+1)}$

Simplify the exponents:

$\frac{1}{3^{3x-2}} = 3^{16x+4}$

Next, we can equate the bases since the exponents must be equal:

$3^{3x-2} = 3^{16x+4}$

Since the bases are the same, we can now set the exponents equal to each other:

$3x-2 = 16x+4$

Now, solve for $x$:

Subtract $3x$ from both sides:

$-2 = 13x + 4$

Subtract $4$ from both sides:

$-6 = 13x$

Finally, divide by $13$ to isolate $x$:

$x = \frac{-6}{13}$

So, the solution to the equation is $x = \frac{-6}{13}$.