# Solve the exponent equation :81^2x = 27^(4x-1).

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### 3 Answers

We have to solve : 81^2x = 27^(4x-1)

81^2x = 27^(4x-1)

Let's make the base equal on both the side

=> 3^4^2x = 3^3^(4x - 1)

=> 3^8x = 3^(12x - 3)

Now we can equate the exponent

8x = 12x - 3

=> 4x = 3

=> x = 3/4

**The solution of the equation is x = 3/4**

81^2x = 27^(4x-1)

First we will simplify by factoring the bases.

==> 81 = 3*3*3*3= 3^4

==> 27 = 3*3*3 = 3^3

Now we will substitute.

==> 3^4^2x = 3^3^(4x-1)

But we know that x^a^b= x^ab.

==> 3^(4*2x) = 3^(3*(4x-1)

==> 3^8x = 3^(12x-3)

Now that the bases are equal, then the powers are equal.

==> 8x = 12x -3

==> -4x = -3

==> x = 3/4

**Then the solution to the exponent equation is x= 3/4**

The equation `81^(2x) = 27^(4x-1)` has to be solved.

First equate the base of both sides of the equation.

`81^(2x) = 27^(4x-1)`

`81 = 3^4` and `27 = 3^3`

This gives:

`(3^4)^(2x) = (3^3)^(4x - 1)`

`3^(8x) = 3^(12x - 3)`

Now, the exponent terms of both the sides can be equated to solve for x.

8x = 12x - 3

4x = 3

x = 3/4

The solution of the equation is x = 3/4.