What is the answer to log_{2}(x)+2log_{4}(x-1)-1=0 ?

The answer to log_{2}(x)+2log_{4}(x-1)-1=0 is x=2

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log_{2}(x)+2log_{4}(x-1)-1=0

lo g_{2} (x) + 2 lo g_{4} (x - 1) - 1 = 0

x = 2

Solution steps

lo g_{2} (x) + 2 lo g_{4} (x - 1) - 1 = 0

Apply log rules

x^{2} (x - 1)^{2} = 4

Solve

x^{2} (x - 1)^{2} = 4 : x = - 1, x = 2

x = - 1, x = 2

Verify Solutions:

x = - 1

False

, x = 2

True

The solution is

x = 2

27 = \frac{75}{ln ( x + 1 )}

lo g_{8} (x) - lo g_{8} (x - 3) = 2

lo g_{2 x - 3} (3 x^{2} - 18) = 2

4 ln (5 x - 4) - 11 = 5

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

What is the answer to log_{2}(x)+2log_{4}(x-1)-1=0 ?
The answer to log_{2}(x)+2log_{4}(x-1)-1=0 is x=2
Solve for x: log_{2}(x)+2log_{4}(x-1)-1=0
The solution is x=2