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

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

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

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

x = 4

Solution steps

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

Add

lo g_{3} (x - 1)

to both sides

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

Simplify

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

Apply log rules

2 x + 1 = 3 (x - 1)

Solve

2 x + 1 = 3 (x - 1) : x = 4

x = 4

Verify Solutions:

x = 4

True

The solution is

x = 4

lo g_{10} (x) = - lo g_{10} (6)

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

15 ln (x + 1) = 75

lo g_{2} (x) + lo g_{3} (5) = 2

lo g_{2} ((x + 13) (5 x + 1)) = 8

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