Solve the equation. log 5 (x + 1) = 1 + log 5 (x - 1)

Question

Solve the equation.
 log₅ (x + 1) = 1 + log₅ (x - 1)

Answer 1

1) Subtract Log₅(x-1) from both sides, so now we have:

Log₅(x+1)-Log₅(x-1) =1

2) Use the Log rule Log_n(a)-Log_n(b) = Log_n(a/b) to combine the two Logs into one Log:

Log₅((x+1)/(x-1))=1

3) "Undo" the Log by putting both sides in the exponent of base 5 (this is how you undo a Log):

5^(Log5((x+1)/(x-1)))=5^1

(x+1)/(x-1)=5

4) Assuming x is not 1 (we can check this later), multiply both sides by (x-1):

x+1=5(x-1)

x+1=5x-5

5) Get all the x's on one side (right), all the other stuff on the left, and solve for x by dividing:

6=4x

6/4=x

3/2=x

Solution: x=3/2.  We can check now that x cannot be 1 because if we plug this into the original problem, the Log on the right blows up (approaches negative infinity). So x=1 can't be a solution, which means step 4 was okay.

Answer 2

Hi Couva,

 

log5 (x + 1) = 1 + log5 (x - 1)

log5 (x + 1) - log5 (x - 1) = 1  

log5((x+1)/(x-1)) = 1 ( I can write the log that way based on quotient identity)

5 = (x+1) / (x-1)

5x - 5 = x + 1

4x = 6

x = 3/2