What is the answer to log_{4}(x+5)-log_{4}(1-5x)=2 ?

The answer to log_{4}(x+5)-log_{4}(1-5x)=2 is x= 11/81

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log_{4}(x+5)-log_{4}(1-5x)=2

lo g_{4} (x + 5) - lo g_{4} (1 - 5 x) = 2

x = \frac{11}{81}

Decimal Notation

x = 0.13580 \dots

Solution steps

lo g_{4} (x + 5) - lo g_{4} (1 - 5 x) = 2

Add

lo g_{4} (1 - 5 x)

to both sides

lo g_{4} (x + 5) - lo g_{4} (1 - 5 x) + lo g_{4} (1 - 5 x) = 2 + lo g_{4} (1 - 5 x)

Simplify

lo g_{4} (x + 5) = 2 + lo g_{4} (1 - 5 x)

Apply log rules

x + 5 = 16 (1 - 5 x)

Solve

x + 5 = 16 (1 - 5 x) : x = \frac{11}{81}

x = \frac{11}{81}

Verify Solutions:

x = \frac{11}{81}

True

The solution is

x = \frac{11}{81}

- ln (m^{2} - m + 1) = 0

p, ln (p) - ln (- p + 1) = t

3 lo g_{a^{3}} (x) = lo g_{a} (x)

3.547 = lo g_{10} (x)

2 - 6 ln (3 x) = 10

What is the answer to log_{4}(x+5)-log_{4}(1-5x)=2 ?
The answer to log_{4}(x+5)-log_{4}(1-5x)=2 is x= 11/81
Solve for x: log_{4}(x+5)-log_{4}(1-5x)=2
The solution is x= 11/81