Question

Solve: `27^(4x) = (1/9)^(x+5)` ?

Answer 1

I tried this:

Explanation:

You have:
`27^(4x)=(1/9)^(x+5)`
We can write:
`(3^3)^(4x)=(1^2/3^2)^(x+5)`
We now need to use some properties of exponents to write:
`3^(12x)=(1/3)^(2x+10)`
`3^(12x)=(3^-1)^(2x+10)`
`3^(12x)=(3)^(-2x-10)`
Now we apply the log in base `3` on both sides:
`log_3(3^(12x))=log_3(3^(-2x-10))`
That for the definition of log leads us to cancel the log and `3` to get:
`12x=-2x-10`
`14x=-10`
`x=-10/14=-5/7`

Answer 2

`x=-5/7`

Explanation:

With questions like this before taking logs you should look to see if you can reduce both sides of the equation to a common base.

`27^(4x) = (1/9)^(x+5)`

Remember that: `(1/a)^(x) = a^(-x)`

So, in our equation:

`27^(4x) = 9^-(x+5)`

Now consider:

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

Hence, `(3^3)^(4x) = (3^2)^-(x+5)`

Remember that: `(m^p)^q = m^(pxxq)`

So, in our simplified equation:

`3^(12x) = 3^(-2(x+5))`

Now that we have both sides of the equation on the same base `(3)` we can equate exponents.

`:. 12x = -2(x+5)`

`12x+2x = -10`

`14x =-10`

`x=-5/7`