What is x if 81^(x-1)>3^2(x+1)?
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calendarEducator since 2008
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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)
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calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
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).
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