how to solve log(x+3)+log(x-3)=log27 ?

log(x+3)+log(x-3)=log27
==>
Note that x+3>0 and x-3>0 ==> x must be greater than 3
log(x+3)(x-3)=log27
==>
x^2-9=27
==>
x^2=36
==> x=6
and x= -6 (is rejected)
thus,
x=6

exchange LHS to 3*log(x) divide the two factors by using log(x) {this ability that log(x)=0 is likewise a answer so x=a million is a answer (10^a million=0)} 3=log(x) x=1000 so the two recommendations are: x=a million and x=1000

log (x + 3) + log (x – 3) = log 27
log [(x + 3)(x – 3)] = log 27
log x² – 3² = log 27
log x² – 9 = log 27
Compare:
x² – 9 = 27
x² = 36
x = ±√36
x = ±6
Since this is log, x > 0
∴ x = 6

Done

log(x+3)+log(x-3)=log27
log( (x+3) * (x-3) ) = log27
(x+3) * (x-3) = 27
x^2 - 9 = 27
x^2 = 27 + 9
x^2 = 36
x = Sqrt( 36 )

x1 = 6
x2 = -6

-6 can't be solution because

log(x-3) = log(-6-3) = log(-9)

But log(w) is defined for w > 0. Therefore solution is

x = 6

{}

log(x + 3) + log(x - 3) = log(27)
log[(x + 3)(x - 3)] = log(27)
log(x^2 + 3x - 3x - 9) = log(27)
log(x^2 - 9) = log(27)
x^2 - 9 = 27
x^2 = 9 + 27
x = √36
x = 6 [-6 cannot be an answer because log(-6 + 3) and log(-6 - 3) are unacceptable]

log [ (x + 3 ) ( x - 3 ) ] = log 27

( x + 3 ) ( x - 3 ) = 27

x ² - 9 = 27

x ² = 36

x = 6 is acceptable

log(x+3)+log(x-3)=log27
=> log(x+3)(x-3)=log27
=> (x+3)(x-3)=27
=>x^2-9=27
=>x^2=36
=>x=6 or -6

x^2-9=27
x=6

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you need to know the log rythems