What is the answer to log_{5}(2x+1)-log_{5}(x-2)=1 ?

The answer to log_{5}(2x+1)-log_{5}(x-2)=1 is x= 11/3

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log_{5}(2x+1)-log_{5}(x-2)=1

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

x = \frac{11}{3}

Solution steps

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

Add

lo g_{5} (x - 2)

to both sides

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

Simplify

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

Apply log rules

2 x + 1 = 5 (x - 2)

Solve

2 x + 1 = 5 (x - 2) : x = \frac{11}{3}

x = \frac{11}{3}

Verify Solutions:

x = \frac{11}{3}

True

The solution is

x = \frac{11}{3}

lo g_{7} (7 x) + 2 lo g_{7} (6) = 3

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

lo g_{3} (10) - lo g_{3} (- 2 x) = 4

lo g_{3} (8 x - 2) + lo g_{3} (x - 1) = 2

lo g_{5} (X) = - 2

What is the answer to log_{5}(2x+1)-log_{5}(x-2)=1 ?
The answer to log_{5}(2x+1)-log_{5}(x-2)=1 is x= 11/3
Solve for x: log_{5}(2x+1)-log_{5}(x-2)=1
The solution is x= 11/3