What is the answer to log_{5}(x+1)=2+log_{5}(x-1) ?

The answer to log_{5}(x+1)=2+log_{5}(x-1) is x= 13/12

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log_{5}(x+1)=2+log_{5}(x-1)

lo g_{5} (x + 1) = 2 + lo g_{5} (x - 1)

x = \frac{13}{12}

Decimal Notation

x = 1.08333 \dots

Solution steps

lo g_{5} (x + 1) = 2 + lo g_{5} (x - 1)

Apply log rules

x + 1 = 25 (x - 1)

Solve

x + 1 = 25 (x - 1) : x = \frac{13}{12}

x = \frac{13}{12}

Verify Solutions:

x = \frac{13}{12}

True

The solution is

x = \frac{13}{12}

e^{2} ln (b) = 400

x = 4 lo g_{10} (x)

lo g_{10} (x) = 9.33

ln (x + 1) - ln (5) = 4

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

What is the answer to log_{5}(x+1)=2+log_{5}(x-1) ?
The answer to log_{5}(x+1)=2+log_{5}(x-1) is x= 13/12
Solve for x: log_{5}(x+1)=2+log_{5}(x-1)
The solution is x= 13/12