how do you solve ln(x-3)+ln(x+4)=3ln2?

ln(x-3)+ln(x+4)=3ln(2)

ln((x-3)(x+4))=ln(2^3)

(x-3)(x+4)=8

x^2 +x -20=0

x^2 +5x -4x -20=0

(x+5)(x-4)=0

x=-5 or x=4


Welcome!

ln [ (x - 3)(x + 4) ] = ln ( 2 ³ )

(x - 3)(x + 4) = 8
x ² + x - 20 = 0
(x + 5)(x - 4) = 0
x = - 5 , x = 4
Accept x = 4

take the exponent of both sides:

exp[ln(x-3) + ln(x+4)] = exp[3ln(2)]

Now use the rule exp(a+b) = exp(a)*exp(b):

exp[ln(x-3)]*exp[ln(x+4)] = exp[3ln(2)]

But exp[ln( )] = ( ) , so we get [x-3]*[x+4] = {exp[ln(2)]}^3

[x-3]*[x+4] = 2^3 = 8

x^2 + x - 12 - 8 = 0

x^2 + x - 20 = 0

(x + 5)*(x - 4) = 0

x = -5 or x = 4

But only the second solution will satisfy the original equation. You cannot take the natural log of a negative number (unless you want to talk about complex numbers). So the answer is: X = 4.

Hope this helps.

GLN77

ln((x-3)*(x+4))=ln(2^3) basic log rules
ln(x^2 + x - 12) = ln (8)
e^[ln(x^2 + x - 12)] = e^[ln (8)]
x^2 + x - 12 = 8
x^2 + x - 20 = 0
(x+5)(X-4)=0
x=-5,4

ln(x - 3) + ln(x + 4) = 3ln2
ln(x - 3)(x + 4) = 3ln2
ln(x^2 - 3x + 4x - 12) = ln(2^3)
x^2 + x - 12 = 8
x^2 + x - 12 - 8 = 0
x^2 + x - 20 = 0
x^2 + 5x - 4x - 20 = 0
(x^2 + 5x) - (4x + 20) = 0
x(x + 5) - 4(x + 5) = 0
(x + 5)(x - 4) = 0

x + 5 = 0
x = -5

x - 4 = 0
x = 4

∴ x = -5 , 4
*but if we substitute -5 into ln(x - 3), we will get ln(-5 - 3) = ln(-8) (imaginary number), so x only equals 4.

ln(x - 3) + ln(x + 4) = 3ln(2)
ln[(x - 3).(x + 4)] = ln[(2)^3]
(x - 3).(x + 4) = (2)^3
x^2 + x -12 = 8
x^2 + x -20 = 0
(x + 5)( x - 4) = 0
therefore x = 4, -5 Answer