Algebra Help please?
1 answer:
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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,
and rewrite it as
We undo natural log by raising e to both sides:
Once this is canceled we are left with:
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,
and rewrite it as
We undo natural log by raising e to both sides:
Once this is canceled we are left with:
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.