Respuesta :
The answers are x=6 or x=-1.
To solve this, we first use the quotient rule for the natural logarithm. It says that ln(x/y) = ln(x) - ln(y). Applying this, we take our original equation,
[tex]\ln(x-3)-\ln(x+3)=\ln(\frac{2}{x})[/tex]
and rewrite it as
[tex]\ln\frac{x-3}{x+3}=\ln\frac{2}{x}[/tex]
We undo natural log by raising e to both sides:
[tex]e^{\ln\frac{x-3}{x+3}}=e^{\ln\frac{2}{x}}[/tex]
Once this is canceled we are left with:
[tex]\frac{x-3}{x+3}=\frac{2}{x}[/tex]
If we view this as a proportion, we can cross multiply:
(x-3)(x) = 2(x+3)
Using the distributive property on both sides we have:
x²-3x=2x+6
Cancel the 2x from the right hand side by subtracting:
x²-3x-2x=2x+6-2x
x²-5x=6
Cancel the 6 by subtracting:
x²-5x-6=0
This is easily factorable. We want factors of -6 that sum to -5; -6(1) = -6 and -6+1=-5:
(x-6)(x+1)=0
Using the zero product property, we know either x-6=0 or x+1=0; thus x=6 or x=-1.
To solve this, we first use the quotient rule for the natural logarithm. It says that ln(x/y) = ln(x) - ln(y). Applying this, we take our original equation,
[tex]\ln(x-3)-\ln(x+3)=\ln(\frac{2}{x})[/tex]
and rewrite it as
[tex]\ln\frac{x-3}{x+3}=\ln\frac{2}{x}[/tex]
We undo natural log by raising e to both sides:
[tex]e^{\ln\frac{x-3}{x+3}}=e^{\ln\frac{2}{x}}[/tex]
Once this is canceled we are left with:
[tex]\frac{x-3}{x+3}=\frac{2}{x}[/tex]
If we view this as a proportion, we can cross multiply:
(x-3)(x) = 2(x+3)
Using the distributive property on both sides we have:
x²-3x=2x+6
Cancel the 2x from the right hand side by subtracting:
x²-3x-2x=2x+6-2x
x²-5x=6
Cancel the 6 by subtracting:
x²-5x-6=0
This is easily factorable. We want factors of -6 that sum to -5; -6(1) = -6 and -6+1=-5:
(x-6)(x+1)=0
Using the zero product property, we know either x-6=0 or x+1=0; thus x=6 or x=-1.
