What is the answer to log_{7}(n-4)+log_{7}(n+2)=1 ?

The answer to log_{7}(n-4)+log_{7}(n+2)=1 is n=5

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log_{7}(n-4)+log_{7}(n+2)=1

lo g_{7} (n - 4) + lo g_{7} (n + 2) = 1

n = 5

Solution steps

lo g_{7} (n - 4) + lo g_{7} (n + 2) = 1

Apply log rules

(n - 4) (n + 2) = 7

Solve

(n - 4) (n + 2) = 7 : n = 5, n = - 3

n = 5, n = - 3

Verify Solutions:

n = 5

True

, n = - 3

False

The solution is

n = 5

lo g_{7} (3 x) = A

ln (x + 8) + ln (x - 4) = 2 ln (x)

(ln (x))^{2} - 1 = 0

lo g_{x} (\frac{25}{9}) = 2

ln (2 - x) = 3

What is the answer to log_{7}(n-4)+log_{7}(n+2)=1 ?
The answer to log_{7}(n-4)+log_{7}(n+2)=1 is n=5
Solve for n: log_{7}(n-4)+log_{7}(n+2)=1
The solution is n=5