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

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

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log_{e}(x+e)-log_{e}(x-e)=1

lo g_{e} (x + e) - lo g_{e} (x - e) = 1

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

Decimal Notation

x = 5.88223 \dots

Solution steps

lo g_{e} (x + e) - lo g_{e} (x - e) = 1

Add

lo g_{e} (x - e)

to both sides

lo g_{e} (x + e) - lo g_{e} (x - e) + lo g_{e} (x - e) = 1 + lo g_{e} (x - e)

Simplify

lo g_{e} (x + e) = 1 + lo g_{e} (x - e)

Apply log rules

x + e = e (x - e)

Solve

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

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

Verify Solutions:

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

True

The solution is

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

3 + ln (x) = 3 ln (x)

\frac{9}{7} lo g_{4} (x) = - 4

lo g_{4} (x) = - 2

1 = 3.5 + lo g_{10} (x)

- lo g_{10} (x) = 1 0^{- 2}

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