What is the answer to log_{9}(x+1)-log_{9}(1-x)=log_{9}(2x+3) ?

The answer to log_{9}(x+1)-log_{9}(1-x)=log_{9}(2x+3) is x=(sqrt(5)-1)/2

Solve for x: log_{9}(x+1)-log_{9}(1-x)=log_{9}(2x+3)

The solution is x=(sqrt(5)-1)/2

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lo g_{9} (x + 1) - lo g_{9} (1 - x) = lo g_{9} (2 x + 3)

x = \frac{5 - 1}{2}

Decimal Notation

x = 0.61803 \dots

Solution steps

lo g_{9} (x + 1) - lo g_{9} (1 - x) = lo g_{9} (2 x + 3)

Add

lo g_{9} (1 - x)

to both sides

lo g_{9} (x + 1) - lo g_{9} (1 - x) + lo g_{9} (1 - x) = lo g_{9} (2 x + 3) + lo g_{9} (1 - x)

Simplify

lo g_{9} (x + 1) = lo g_{9} (2 x + 3) + lo g_{9} (1 - x)

Apply log rules

x + 1 = (2 x + 3) (1 - x)

Solve

x + 1 = (2 x + 3) (1 - x) : x = - \frac{1 + 5}{2}, x = \frac{5 - 1}{2}

x = - \frac{1 + 5}{2}, x = \frac{5 - 1}{2}

Verify Solutions:

x = - \frac{1 + 5}{2}

False

, x = \frac{5 - 1}{2}

True

The solution is

x = \frac{5 - 1}{2}

u, ln (1 - u) = x

ln (ln (x)) = 1

2 \cdot ln (x) + ln (x^{2} + 2) = ln (3)

lo g_{5} (x - 3) = 2

lo g_{2} (3 x - 1) - lo g_{2} (x + 2) = 4

What is the answer to log_{9}(x+1)-log_{9}(1-x)=log_{9}(2x+3) ?
The answer to log_{9}(x+1)-log_{9}(1-x)=log_{9}(2x+3) is x=(sqrt(5)-1)/2
Solve for x: log_{9}(x+1)-log_{9}(1-x)=log_{9}(2x+3)
The solution is x=(sqrt(5)-1)/2