And another logarithm

[Koala]

Okay, this one was on my last test, and it still kills me. I need help understanding not just how we do this, but why we do the steps we do.
log₈(x+6)=1-log₈(x+4)
Whoever explains this one, please go super slow and tell me why you do what you do. Thanks so much!

 

[pappus]

Okay, this one was on my last test, and it still kills me. I need help understanding not just how we do this, but why we do the steps we do.
log₈(x+6)=1-log₈(x+4)
Whoever explains this one, please go super slow and tell me why you do what you do. Thanks so much!

1. Express \(\displaystyle 1 = \log_8(8)\). Use the logarithmic law: \(\displaystyle \log\left(\frac ab\right) = \log(a)-\log(b)\)

2. Your equation becomes now:

\(\displaystyle \log_8(x+6) = \log_8\left(\frac8{x+4}\right)\)

3. Since the logarithms have the same base the arguments must be equal too:

\(\displaystyle x+6 = \frac8{x+4}~\implies~x^2+10x+16=0\) yields \(\displaystyle x = -8~\vee~x=-2\)

4. Obviously x = -8 is not a valid solution because you would get a negative argument. So x = -2. Confirmation:

\(\displaystyle \log_8((-2)+6) = 1-\log_8((-2)+4)~\implies~\frac23 = 1-\frac13\) which is obviously true.

 

[Koala]

Awesome, thanks! I think I get whatever it was I was missing before.