The formula for dB in voltage (or current) is 20*log(V2/V1). Here that's 20*log(5/10)=-6dB. This is the value you can always expect when the output voltage or current is half of the input. The 20 comes from 2*10. The 10 is there because these are deci-bels and not bels. The 2 is there because the decibel is a measure of ratios of powers. Power in electric is proportional to the square of voltage or the square of current. Because of the rules of logarithms, the 2 exponent of the square is brought out front in the equation as a multiplier. Therefore in comparing output Power to input Power given in watts, you don't need the 2, and use only the formula 10*log(P2/P1). For half power its -3dB. But for your answer its -6dB
