What is the answer to log_{a}(-x)-log_{a}(2-x)=2 ?

The answer to log_{a}(-x)-log_{a}(2-x)=2 is x=(2a^2)/(-1+a^2) {0<a<1}

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log_{a}(-x)-log_{a}(2-x)=2

lo g_{a} (- x) - lo g_{a} (2 - x) = 2

x = \frac{2 a ^{2}}{- 1 + a ^{2}} {0 < a < 1}

Solution steps

lo g_{a} (- x) - lo g_{a} (2 - x) = 2

Add

lo g_{a} (2 - x)

to both sides

lo g_{a} (- x) - lo g_{a} (2 - x) + lo g_{a} (2 - x) = 2 + lo g_{a} (2 - x)

Simplify

lo g_{a} (- x) = 2 + lo g_{a} (2 - x)

Apply log rules

- x = a^{2} (2 - x)

Solve

- x = a^{2} (2 - x) : x = \frac{2 a ^{2}}{- 1 + a ^{2}}

x = \frac{2 a ^{2}}{- 1 + a ^{2}}

Verify Solutions:

x = \frac{2 a ^{2}}{- 1 + a ^{2}} {0 < a < 1}

The solution is

x = \frac{2 a ^{2}}{- 1 + a ^{2}} {0 < a < 1}

ln (x) = 5.77

ln (x - 8) = ln (x^{3} + x)

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

x, ln (x^{2}) = 5

2 lo g_{10} (z) - lo g_{10} (z + 7) = 1

What is the answer to log_{a}(-x)-log_{a}(2-x)=2 ?
The answer to log_{a}(-x)-log_{a}(2-x)=2 is x=(2a^2)/(-1+a^2) {0<a<1}