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

The answer to log_{6}(x-2)+log_{6}(x-1)=1 is x=4

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

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

x = 4

Solution steps

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

Apply log rules

(x - 2) (x - 1) = 6

Solve

(x - 2) (x - 1) = 6 : x = 4, x = - 1

x = 4, x = - 1

Verify Solutions:

x = 4

True

, x = - 1

False

The solution is

x = 4

y, - 1 (ln (y)) = ln (x) + c

y, - ln (4 - y) = x

lo g_{10} (2 x) - 20 = 2

2.34 = ln (x)

ln (z) - ln (3) = \frac{3}{2}

What is the answer to log_{6}(x-2)+log_{6}(x-1)=1 ?
The answer to log_{6}(x-2)+log_{6}(x-1)=1 is x=4
Solve for x: log_{6}(x-2)+log_{6}(x-1)=1
The solution is x=4