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ln(x+1)-ln(x-2)=ln(5)

1. Solve equation by setting up algebraically.
2. Use appropriate properties of Logarithm to reduce the number of logarithmic terms to only one.
3. Solve logarithmic equation.

ln(x + 1) - ln(x + 2) = ln5,

2. ln((x + 1) / (x + 2))) = ln5.

3. ln((x + 1) / (x + 2)) = 1.6094,
(x + 1) / (x + 2) = e^1.6094,
(x + 1) / (x + 2) = 5,
Cross multiply:
5X + 10 = X + 1,
5X - X = 1 - 10,
4X = - 9,
X = -2 1/4 = -9/4.

CORRECTION:I copied the problem wrong.

ln(x + 1) - ln(x - 2) = ln5.

2. ln((x + 1) / (x - 2)) = ln5.

3. ln((x + 1) / (x - 2)) = 1.6094,
(x + 1) / (x - 2) = e^1.6094,
(x + 1) / (x - 2) = 5,
Cross multiply:
5X - 10 = X + 1,
5X - X = 1 + 10,
4X = 11,
X = 2 3/4 = 11/4.

3lnx+2lny-4lnz

Sure! I'd be happy to help you with this equation.

1. Solve equation by setting up algebraically:
Start by applying the property of logarithms that states: ln(a) - ln(b) = ln(a/b). We can rewrite the equation as follows:
ln((x+1)/(x-2)) = ln(5)

2. Use appropriate properties of logarithm to reduce the number of logarithmic terms to only one:
Since we have logarithms on both sides of the equation, we can eliminate the logarithmic terms by taking the exponent of both sides. The property ln(a) = b is equivalent to e^b = a (where e is the base of the natural logarithm).
Taking the exponent of both sides of the equation, we get:
e^(ln((x+1)/(x-2))) = e^(ln(5))
Simplifying the left side, we have:
(x+1)/(x-2) = 5

3. Solve logarithmic equation:
To solve this equation, we will isolate the variable x. Multiply both sides of the equation by (x-2) to get rid of the denominator:
(x+1) = 5(x-2)
Expand and simplify:
x + 1 = 5x - 10

Now, your goal is to isolate x on one side of the equation. Subtract x from both sides to collect all the x terms on one side:
1 = 4x - 10

Next, add 10 to both sides:
11 = 4x

Finally, divide both sides of the equation by 4 to solve for x:
x = 11/4

Therefore, the solution to the equation ln(x+1)-ln(x-2)=ln(5) is x = 11/4.

Remember to always check your solution by plugging it back into the original equation to ensure it satisfies the equation.