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

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

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

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

x = 14

Solution steps

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

Add

lo g_{4} (x - 3)

to both sides

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

Simplify

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

Apply log rules

3 x + 2 = 4 (x - 3)

Solve

3 x + 2 = 4 (x - 3) : x = 14

x = 14

Verify Solutions:

x = 14

True

The solution is

x = 14

2 lo g_{10} (x - 1) = 0

ln (x^{2}) = 0.7

lo g_{2} (x) = 8

lo g_{5} (x) = - 2

ln (x - 1) + ln (x + 1) = 0

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