What is the answer to log_{7}(3x+2)-log_{7}(x-2)=1 ?

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

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log_{7}(3x+2)-log_{7}(x-2)=1

lo g_{7} (3 x + 2) - lo g_{7} (x - 2) = 1

x = 4

Solution steps

lo g_{7} (3 x + 2) - lo g_{7} (x - 2) = 1

Add

lo g_{7} (x - 2)

to both sides

lo g_{7} (3 x + 2) - lo g_{7} (x - 2) + lo g_{7} (x - 2) = 1 + lo g_{7} (x - 2)

Simplify

lo g_{7} (3 x + 2) = 1 + lo g_{7} (x - 2)

Apply log rules

3 x + 2 = 7 (x - 2)

Solve

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

x = 4

Verify Solutions:

x = 4

True

The solution is

x = 4

lo g_{b} (128) = 7

2 - ln (7 - x) = 0

lo g_{x - 1} (145) = 7

ln (2 x + 1) = - 4

ln (x) = 3.9

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