What is the answer to log_{7}(12x+1)+log_{7}(x)=0 ?

The answer to log_{7}(12x+1)+log_{7}(x)=0 is x= 1/4

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log_{7}(12x+1)+log_{7}(x)=0

lo g_{7} (12 x + 1) + lo g_{7} (x) = 0

x = \frac{1}{4}

Decimal Notation

x = 0.25

Solution steps

lo g_{7} (12 x + 1) + lo g_{7} (x) = 0

Apply log rules

(12 x + 1) x = 1

Solve

(12 x + 1) x = 1 : x = \frac{1}{4}, x = - \frac{1}{3}

x = \frac{1}{4}, x = - \frac{1}{3}

Verify Solutions:

x = \frac{1}{4}

True

, x = - \frac{1}{3}

False

The solution is

x = \frac{1}{4}

lo g_{4} (x + 17) + lo g_{4} (x + 5) = 3

ln (x) - 3 ln (x) + 2 = 0

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

ln (2 x + 4) = ln (x + 3)

lo g_{3} (x + 12) = lo g_{3} (5 x)

What is the answer to log_{7}(12x+1)+log_{7}(x)=0 ?
The answer to log_{7}(12x+1)+log_{7}(x)=0 is x= 1/4
Solve for x: log_{7}(12x+1)+log_{7}(x)=0
The solution is x= 1/4