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

The answer to log_{7}(x+4)-log_{7}(x-2)=1 is x=3

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

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

x = 3

Solution steps

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

Add

lo g_{7} (x - 2)

to both sides

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

Simplify

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

Apply log rules

x + 4 = 7 (x - 2)

Solve

x + 4 = 7 (x - 2) : x = 3

x = 3

Verify Solutions:

x = 3

True

The solution is

x = 3

0 = 2 lo g_{10} (x)

ln (x^{2} - 5 x - 1) - ln (3 - 2 x) = 0

ln (\frac{x}{x ^{2} + 1}) = 1

lo g_{2} (x) + 3 lo g_{10} (x) + 2 = 0

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

What is the answer to log_{7}(x+4)-log_{7}(x-2)=1 ?
The answer to log_{7}(x+4)-log_{7}(x-2)=1 is x=3
Solve for x: log_{7}(x+4)-log_{7}(x-2)=1
The solution is x=3