What is the answer to log_{6}(3x+2)+log_{6}(x-5)=1 ?

The answer to log_{6}(3x+2)+log_{6}(x-5)=1 is x= 16/3

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log_{6}(3x+2)+log_{6}(x-5)=1

lo g_{6} (3 x + 2) + lo g_{6} (x - 5) = 1

x = \frac{16}{3}

Decimal

x = 5.33333 \dots

Solution steps

lo g_{6} (3 x + 2) + lo g_{6} (x - 5) = 1

Apply log rules

(3 x + 2) (x - 5) = 6

Solve

(3 x + 2) (x - 5) = 6 : x = \frac{16}{3}, x = - 1

x = \frac{16}{3}, x = - 1

Verify Solutions:

x = \frac{16}{3}

True

, x = - 1

False

The solution is

x = \frac{16}{3}

- 1.2 = lo g_{10} (x)

lo g_{5} (x) = 9 lo g_{x} (5)

lo g_{2} (x + 6) + lo g_{2} (x + 7) = 1

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

lo g_{2} (n) = 4

What is the answer to log_{6}(3x+2)+log_{6}(x-5)=1 ?
The answer to log_{6}(3x+2)+log_{6}(x-5)=1 is x= 16/3
Solve for x: log_{6}(3x+2)+log_{6}(x-5)=1
The solution is x= 16/3