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

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)

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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

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