What is the answer to log_{2}(x+1)+log_{2}(x-1)=4 ?

The answer to log_{2}(x+1)+log_{2}(x-1)=4 is x=sqrt(17)

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log_{2}(x+1)+log_{2}(x-1)=4

lo g_{2} (x + 1) + lo g_{2} (x - 1) = 4

x = 17

Decimal Notation

x = 4.12310 \dots

Solution steps

lo g_{2} (x + 1) + lo g_{2} (x - 1) = 4

Apply log rules

(x + 1) (x - 1) = 16

Solve

(x + 1) (x - 1) = 16 : x = 17, x = - 17

x = 17, x = - 17

Verify Solutions:

x = 17

True

, x = - 17

False

The solution is

x = 17

ln (\frac{( t - 63 )}{125}) = - 0.07 t

2 ln (4 x + 1) + 14 = 20

7.5 = - lo g_{10} (h)

x, lo g_{10} (x) = - 0.123

- lo g_{10} (x) = 12

What is the answer to log_{2}(x+1)+log_{2}(x-1)=4 ?
The answer to log_{2}(x+1)+log_{2}(x-1)=4 is x=sqrt(17)
Solve for x: log_{2}(x+1)+log_{2}(x-1)=4
The solution is x=sqrt(17)