solve the equation log2 (x-1)=2-log2 (x+1) those 2s behind the log are the base

Question

^{thank you s o m u c h p l e a s e}

Answer 1

Hi Thomas,

 

 

Log₂(x-1) = 2 - log₂(x+1)

 

isolate the logs, collect on left side

 

log₂(x-1) + log₂(x+1) = 2

 

apply log property, sum of logs is equal to log of products

 

log₂(x-1)(x+1) = 2

 

Multiply two binomials FOIL 

 

log₂(x²-1) = 2

 

undo log by exponentiating both sides with a base of 2, or by log definition, the base 2 raised to a power of 2 is equal to (x²-1)

 

2^{2 =} (x²-1)  simplify then solve for x

 

4 = x² - 1

 

4 + 1 = x²

x²  = 5. take sqrt of both sides to undo squaring

x  = + or - sqrt of 5

 

check solution, remember you cannot take log of zero or negative number

 

f you plug in -sqrt of 5 on the left log of the original equation, you will have to take a log of negative number so  -sqrt of 5 is not a solution.

 

x = sqrt 5

 

Hope it helps

 

 

 

 

Answer 2

Teresa G.'s solution pointed out my error.

A revision:

 

log₂(x - 1) = 2 - log₂(x + 1)

log₂(x - 1) + log₂(x + 1) = 2  add log₂(x - 1) to both sides

log₂[(x - 1)(x + 1)] = 2          sum/product rule

log₂(x² - 1) = 2                      difference of square

2² = x² - 1                             exponential form

4 = x² - 1

5 = x²

√5 = x                                  log argument cannot be negative