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

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

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

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

x = \frac{4}{7}

Solution steps

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

Add

lo g_{4} (2 x - 1)

to both sides

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

Simplify

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

Apply log rules

x = 4 (2 x - 1)

Solve

x = 4 (2 x - 1) : x = \frac{4}{7}

x = \frac{4}{7}

Verify Solutions:

x = \frac{4}{7}

True

The solution is

x = \frac{4}{7}

75 = 20 lo g_{10} (\frac{800}{x})

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

6 + 7 \cdot lo g_{2} (x) = 21

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

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

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