What is the answer to log_{3}(3x+1)-log_{3}(x-7)=4 ?

The answer to log_{3}(3x+1)-log_{3}(x-7)=4 is x= 284/39

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log_{3}(3x+1)-log_{3}(x-7)=4

lo g_{3} (3 x + 1) - lo g_{3} (x - 7) = 4

x = \frac{284}{39}

Solution steps

lo g_{3} (3 x + 1) - lo g_{3} (x - 7) = 4

Add

lo g_{3} (x - 7)

to both sides

lo g_{3} (3 x + 1) - lo g_{3} (x - 7) + lo g_{3} (x - 7) = 4 + lo g_{3} (x - 7)

Simplify

lo g_{3} (3 x + 1) = 4 + lo g_{3} (x - 7)

Apply log rules

3 x + 1 = 81 (x - 7)

Solve

3 x + 1 = 81 (x - 7) : x = \frac{284}{39}

x = \frac{284}{39}

Verify Solutions:

x = \frac{284}{39}

True

The solution is

x = \frac{284}{39}

lo g_{10} (y) = 4

lo g_{10} (y) = 9

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

100 = 10 lo g_{10} (x)

ln (\frac{4 x + 1}{3}) = 2

What is the answer to log_{3}(3x+1)-log_{3}(x-7)=4 ?
The answer to log_{3}(3x+1)-log_{3}(x-7)=4 is x= 284/39
Solve for x: log_{3}(3x+1)-log_{3}(x-7)=4
The solution is x= 284/39