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

The answer to log_{2}(x+6)-log_{2}(x-1)=1 is x=8

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

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

x = 8

Solution steps

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

Add

lo g_{2} (x - 1)

to both sides

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

Simplify

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

Apply log rules

x + 6 = 2 (x - 1)

Solve

x + 6 = 2 (x - 1) : x = 8

x = 8

Verify Solutions:

x = 8

True

The solution is

x = 8

lo g_{3} (- 2 x + 15) = 2

lo g_{2} (x - 6) = 7

lo g_{10} (4) = 2 - lo g_{10} (x + 5)

0.23 = lo g_{10} (x)

x, 2 ln (x) = ln (y)

What is the answer to log_{2}(x+6)-log_{2}(x-1)=1 ?
The answer to log_{2}(x+6)-log_{2}(x-1)=1 is x=8
Solve for x: log_{2}(x+6)-log_{2}(x-1)=1
The solution is x=8