What is the answer to log_{5}(x+4)+log_{5}(x-4)=2 ?

The answer to log_{5}(x+4)+log_{5}(x-4)=2 is x=sqrt(41)

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log_{5}(x+4)+log_{5}(x-4)=2

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

x = 41

Decimal Notation

x = 6.40312 \dots

Solution steps

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

Apply log rules

(x + 4) (x - 4) = 25

Solve

(x + 4) (x - 4) = 25 : x = 41, x = - 41

x = 41, x = - 41

Verify Solutions:

x = 41

True

, x = - 41

False

The solution is

x = 41

ln (7 x + 32) + ln (4) = 2 ln (x)

1 = lo g_{10} (2 x^{2} + 3 x - 22)

lo g_{x - 8} (x^{2} - 16) = 2

lo g_{1} (x) = 1.755

lo g_{10} (3 x - 3) = 3

What is the answer to log_{5}(x+4)+log_{5}(x-4)=2 ?
The answer to log_{5}(x+4)+log_{5}(x-4)=2 is x=sqrt(41)
Solve for x: log_{5}(x+4)+log_{5}(x-4)=2
The solution is x=sqrt(41)