# What is x if 81^(x-1)>3^2(x+1)?

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Given the inequality:

81^(x-1) > 3^2(x+1)

To solve the exponent equation, we will rewrite the numbers so the bases are equal.

We know that 81 = 3^4

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

Now, from exponent properties, we know that a^x^y = a^xy

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

Now that the bases are equal, then, the powers should hold the inequality.

==> 4(x-1) > 2(x+1)

We will divide by 2.

==> 2(x-1) > (x+1)

==> 2x -2 > x+ 1

==> x > 3

**Then, x belong to the interval ( 3, inf)**

We need to find x if 81^(x-1)>3^2(x+1)

81^(x-1)>3^2(x+1)

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

As the base is positive and greater than 1, we can write

4(x - 1) > 2(x + 1)

=> 4x - 4 > 2x + 2

=> 2x > 6

=> x > 3

Therefore** x > 3**

We'll write both bases as power of 3:

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

The exponential function is increasing, since the base is bigger than 1.

4(x-1)> 2(x+1)

We'll divide by 2:

2(x- 1) > x + 1

We'll remove the brackets:

2x - 2 > x + 1

We'll subtract x both sides:

2x - x - 2 > 1

x - 2 > 1

We'll add 2 both sides:

x > 2 + 1

x > 3

**The interval of values of x, for the inequality to hold, is (3 , +infinite).**