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

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

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

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

x = 3

Solution steps

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

Apply log rules

5 + x = (x - 1)^{2} \cdot 2

Solve

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

x = 3, x = - \frac{1}{2}

Verify Solutions:

x = 3

True

, x = - \frac{1}{2}

False

The solution is

x = 3

lo g_{10} ((x - 1)^{2}) = 0

x, lo g_{k} (x) = 10

lo g_{7} (x) = - 3

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

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

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