What is the answer to log_{4}(5x+7)-log_{4}(x-3)=2 ?

The answer to log_{4}(5x+7)-log_{4}(x-3)=2 is x=5

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log_{4}(5x+7)-log_{4}(x-3)=2

lo g_{4} (5 x + 7) - lo g_{4} (x - 3) = 2

x = 5

Solution steps

lo g_{4} (5 x + 7) - lo g_{4} (x - 3) = 2

Add

lo g_{4} (x - 3)

to both sides

lo g_{4} (5 x + 7) - lo g_{4} (x - 3) + lo g_{4} (x - 3) = 2 + lo g_{4} (x - 3)

Simplify

lo g_{4} (5 x + 7) = 2 + lo g_{4} (x - 3)

Apply log rules

5 x + 7 = 16 (x - 3)

Solve

5 x + 7 = 16 (x - 3) : x = 5

x = 5

Verify Solutions:

x = 5

True

The solution is

x = 5

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

lo g_{3} (x) + 2 = lo g_{3} (18)

0.027 \cdot 60 = - ln (1 - x)

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

e^{2} ln (b) = 256

What is the answer to log_{4}(5x+7)-log_{4}(x-3)=2 ?
The answer to log_{4}(5x+7)-log_{4}(x-3)=2 is x=5
Solve for x: log_{4}(5x+7)-log_{4}(x-3)=2
The solution is x=5