What is the answer to log_{2}(3x-2)-log_{2}(x-1)=4 ?

The answer to log_{2}(3x-2)-log_{2}(x-1)=4 is x= 14/13

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log_{2}(3x-2)-log_{2}(x-1)=4

lo g_{2} (3 x - 2) - lo g_{2} (x - 1) = 4

x = \frac{14}{13}

Decimal Notation

x = 1.07692 \dots

Solution steps

lo g_{2} (3 x - 2) - lo g_{2} (x - 1) = 4

Add

lo g_{2} (x - 1)

to both sides

lo g_{2} (3 x - 2) - lo g_{2} (x - 1) + lo g_{2} (x - 1) = 4 + lo g_{2} (x - 1)

Simplify

lo g_{2} (3 x - 2) = 4 + lo g_{2} (x - 1)

Apply log rules

3 x - 2 = 16 (x - 1)

Solve

3 x - 2 = 16 (x - 1) : x = \frac{14}{13}

x = \frac{14}{13}

Verify Solutions:

x = \frac{14}{13}

True

The solution is

x = \frac{14}{13}

∣ lo g_{2} (x) ∣ = 3

lo g_{2} (x^{2} - 9) = 4

x, lo g_{e} (2 x + 1) = 0

2 lo g_{10} (x - 2) - lo g_{10} (x) = 0

lo g_{10} (x) 64 = 3

What is the answer to log_{2}(3x-2)-log_{2}(x-1)=4 ?
The answer to log_{2}(3x-2)-log_{2}(x-1)=4 is x= 14/13
Solve for x: log_{2}(3x-2)-log_{2}(x-1)=4
The solution is x= 14/13