What is the answer to log_{2}(x+4)-log_{2}(x-1)=1 ?

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

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log_{2}(x+4)-log_{2}(x-1)=1

lo g_{2} (x + 4) - lo g_{2} (x - 1) = 1

x = 6

Solution steps

lo g_{2} (x + 4) - lo g_{2} (x - 1) = 1

Add

lo g_{2} (x - 1)

to both sides

lo g_{2} (x + 4) - lo g_{2} (x - 1) + lo g_{2} (x - 1) = 1 + lo g_{2} (x - 1)

Simplify

lo g_{2} (x + 4) = 1 + lo g_{2} (x - 1)

Apply log rules

x + 4 = 2 (x - 1)

Solve

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

x = 6

Verify Solutions:

x = 6

True

The solution is

x = 6

lo g_{3} (y) + 4 lo g_{y} (3) = 4

lo g_{\frac{1}{2}} (x) = - 1

lo g_{10} (4 x) + lo g_{10} (8 x) = 1

lo g_{2} (n) = 3.6 \cdot 1 0^{13}

ln (x - 1) + ln (x + 1) = ln (3)

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