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

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

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

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

x = - 6

Solution steps

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

Apply log rules

(x^{2} - 4) \cdot 2 = (2 - x)^{2}

Solve

(x^{2} - 4) \cdot 2 = (2 - x)^{2} : x = - 6, x = 2

x = - 6, x = 2

Verify Solutions:

x = - 6

True

, x = 2

False

The solution is

x = - 6

lo g_{4} (2^{x} + 48) = x - 1

lo g_{4} (x^{2} + 3 x - 1) = lo g_{4} (5 x - 1)

lo g_{x} (126) = 3

ln (2 x) = ln (x + 2)

lo g_{10} (10 x) = 3

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