What is the answer to log_{4}(x+1)+log_{4}(x-1)=4 ?

The answer to log_{4}(x+1)+log_{4}(x-1)=4 is x=sqrt(257)

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log_{4}(x+1)+log_{4}(x-1)=4

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

x = 257

Decimal Notation

x = 16.03121 \dots

Solution steps

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

Apply log rules

(x + 1) (x - 1) = 256

Solve

(x + 1) (x - 1) = 256 : x = 257, x = - 257

x = 257, x = - 257

Verify Solutions:

x = 257

True

, x = - 257

False

The solution is

x = 257

lo g_{5} (b) = - 3

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

lo g_{2} (39 x) - lo g_{2} (x - 5) = 4

2 lo g_{10} (x - 1) = 2 x - 5

lo g_{x} (8 x - 3) - lo g_{x} (4) = 2

What is the answer to log_{4}(x+1)+log_{4}(x-1)=4 ?
The answer to log_{4}(x+1)+log_{4}(x-1)=4 is x=sqrt(257)
Solve for x: log_{4}(x+1)+log_{4}(x-1)=4
The solution is x=sqrt(257)