How do you solve log3(a-3)=1+log3(a+1)

Question

I don't understand how to do it and check it

Answer 1

log₃(a-3)=1+log₃(a+1)

 

a-3>0, a>3; a+1>0, a>-1

-1<3<a

 

log3(a-3)–log3(a+1)=1

 

log3((a-3)/(a+1))=1

 

(a-3)/(a+1) = 3^1

 

a-3=3(a+1)=3a+3

 

-6=2a

 

a=-3; but "a" has to be GREATER THAN 3.

 

So there is NO SOLUTION.

Answer 2

log₃(a-3) = 1 + log₃(a+1)

 

Combining like terms together,

 

log3(a-3) - log3(a+1) = 1

 

log₃ (a-3)   =  1                   [Because, by logarithmic laws, log a - log b = log (a/b) ]

      (a+1)

 

a-3       =  3^1                    [ Because, if  log_a b = p, then b = p^a ]

a+1 

Cross-multiply,

 

a-3 = 3a+3

 

Solve for a,

 

a-3a = 3+3

-2a = 6

 

a = -2