Question

How do you solve `1/2log_a(x + 2) + 1/2log_a(x - 1) = 2/3log_a27`?

Answer 1

I found: `x=(-1+3sqrt(37))/2`

Explanation:

We can try some manipulations:
collect `1/2`:
`1/2[log_a(x+2)+log_a(x-1)]=2/3log_a27`
`log_a(x+2)+log_a(x-1)=2*2/3log_a27`

use the property of logs that says:
`logx+logy=log(xy)`
`log_a[(x+2)(x-1)]=4/3log_a27`

use the other property of logs that says:
`alogx=logx^a`
`log_a[(x+2)(x-1)]=log_a27^(4/3)`

if the two logs must be equal then the argumens must be as well:
`(x+2)(x-1)=27^(4/3)`
`(x+2)(x-1)=root3(27^4)`
`(x+2)(x-1)=81`
`x^2-x+2x-2-81=0`
`x^2+x-83=0`

You can now use the Quadratic Formula to get:
`x_(1,2)=(-1+-sqrt(1+332))/2=`
two solutions:
`x_1=(-1-3sqrt(37))/2` NO it will give you a negative log.
`x_2=(-1+3sqrt(37))/2`