Question

How do you solve the exponential inequality `27^(4x-1)>=9^(3x+8)`?

Answer 1

`x>=9.5/3`

Explanation:

`27^(4x-1)>=9^(3x+8)`

law of indices:
`(a^m*a^n=a^(m+n))`

`27 = 3^3`
`9=3^2`

`therefore 27 = 9^(3/2)`

`27^(4x-1) = 9^(1.5(4x-1))`

`=9^(6x-1.5)`

`9^(6x-1.5)>=9^(3x+8)`

`6x-1.5>=3x+8`

add `1.5`:

`6x>=x+9.5`

subtract `3x`:

`3x>=9.5`

`x>=9.5/3`

Answer 2

`x=19/6`

Explanation:

It would be very nice if we had the same base. Recall that

`27=3^3` and `9=3^2`

With this in mind, we can rewrite our inequality as

`3^(3(4x-1))=3^(2(3x+8)`

We can further simplify the exponents to get

`3^(12x-3)=3^(6x+16)`

Now that our bases are the same, we can equate the exponents:

`12x-3=6x+16`

`6x=19`

`x=19/6`

Hope this helps!