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

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

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

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

x = 2

Solution steps

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

Apply log rules

\frac{ln ( x + 1 )}{ln ( x - 1 )} = - 1

Multiply both sides by

ln (x - 1)

\frac{ln ( x + 1 )}{ln ( x - 1 )} ln (x - 1) = - 1 \cdot ln (x - 1)

Simplify

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

Add

ln (x - 1)

to both sides

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

Simplify

ln (x + 1) + ln (x - 1) = 0

Apply log rules

(x + 1) (x - 1) = 1

Solve

(x + 1) (x - 1) = 1 : x = 2, x = - 2

x = 2, x = - 2

Verify Solutions:

x = 2

True

, x = - 2

False

The solution is

x = 2

lo g_{\frac{2}{5}} (x) = 0

6 - 3 lo g_{10} (2 n) = 0

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

- lo g_{10} (x) = 3.3

20 = lo g_{a} (0.25)

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