What is the answer to log_{10}(10x)-log_{10}(4-x)=2 ?

The answer to log_{10}(10x)-log_{10}(4-x)=2 is x= 40/11

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log_{10}(10x)-log_{10}(4-x)=2

lo g_{10} (10 x) - lo g_{10} (4 - x) = 2

x = \frac{40}{11}

Decimal Notation

x = 3.63636 \dots

Solution steps

lo g_{10} (10 x) - lo g_{10} (4 - x) = 2

Add

lo g_{10} (4 - x)

to both sides

lo g_{10} (10 x) - lo g_{10} (4 - x) + lo g_{10} (4 - x) = 2 + lo g_{10} (4 - x)

Simplify

lo g_{10} (10 x) = 2 + lo g_{10} (4 - x)

Apply log rules

10 x = 100 (4 - x)

Solve

10 x = 100 (4 - x) : x = \frac{40}{11}

x = \frac{40}{11}

Verify Solutions:

x = \frac{40}{11}

True

The solution is

x = \frac{40}{11}

lo g_{4} (y) + 2 = 3 \cdot lo g_{4} (4 y)

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

lo g_{3} (4 x^{3} + 14 x^{2} + 13 x + 3) = lo g_{3} (x + 1)

lo g_{10} (8 x) = \frac{7}{3}

ln (x + 5) - ln (x) = 3

What is the answer to log_{10}(10x)-log_{10}(4-x)=2 ?
The answer to log_{10}(10x)-log_{10}(4-x)=2 is x= 40/11
Solve for x: log_{10}(10x)-log_{10}(4-x)=2
The solution is x= 40/11