What is the answer to log_{7}(x)+log_{7}(6x-1)=1 ?

The answer to log_{7}(x)+log_{7}(6x-1)=1 is x= 7/6

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log_{7}(x)+log_{7}(6x-1)=1

lo g_{7} (x) + lo g_{7} (6 x - 1) = 1

x = \frac{7}{6}

Decimal Notation

x = 1.16666 \dots

Solution steps

lo g_{7} (x) + lo g_{7} (6 x - 1) = 1

Apply log rules

x (6 x - 1) = 7

Solve

x (6 x - 1) = 7 : x = \frac{7}{6}, x = - 1

x = \frac{7}{6}, x = - 1

Verify Solutions:

x = \frac{7}{6}

True

, x = - 1

False

The solution is

x = \frac{7}{6}

lo g_{10} (X + 5) = lo g_{10} (X + 2)

ln (x) = ln (\frac{e}{x})

lo g_{2} (x) = - 5.567133704

30 = - 7020 ln (q)

2 lo g_{3} (x + 2) = lo g_{3} (729)

What is the answer to log_{7}(x)+log_{7}(6x-1)=1 ?
The answer to log_{7}(x)+log_{7}(6x-1)=1 is x= 7/6
Solve for x: log_{7}(x)+log_{7}(6x-1)=1
The solution is x= 7/6