What is the answer to solvefor x,log_{6}(x)+log_{6}(y)=1 ?

The answer to solvefor x,log_{6}(x)+log_{6}(y)=1 is x= 6/y {y>0}

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solvefor x,log_{6}(x)+log_{6}(y)=1

solve for

x, lo g_{6} (x) + lo g_{6} (y) = 1

x = \frac{6}{y} {y > 0}

Solution steps

lo g_{6} (x) + lo g_{6} (y) = 1

Move

lo g_{6} (y)

to the right side

lo g_{6} (x) = 1 - lo g_{6} (y)

Apply log rules

x = \frac{6}{y}

Verify Solutions:

x = \frac{6}{y} {y > 0}

The solution is

x = \frac{6}{y} {y > 0}

lo g_{2} (x - 4) = 0

lo g_{3} (2 x) = 2

a, lo g_{3} (a) - 2 \cdot lo g_{10} (b) = 3

ln (\frac{( x + 1 )}{x}) + ln (2) = ln (x + 3)

lo g_{2} (lo g_{4} (lo g_{5} (x))) = 0

What is the answer to solvefor x,log_{6}(x)+log_{6}(y)=1 ?
The answer to solvefor x,log_{6}(x)+log_{6}(y)=1 is x= 6/y {y>0}