Solve for x? log_4 x - log_4 (x-1) = 1/2?

log4x - log4(x-1) =1/2,

or, log[4x/4(x-1)] =1/2

or, log[x/(x-1)]=1/2

or, log[x/(x-1)]=(1/2)log_e

or, x/(x-1)= e^(1/2)

or, x=xe^1/2 - e^1/2

or, x(1 - e^1/2)=e^1/2

or, x=(e^1/2) / (1 - e^1/2)

Use the exchange of Base theorem. log_4 (x) - log_16 (x+3) = a million/2 so, log_4(x) - log_4(x+3)/ log_4(sixteen) =a million/2 via fact that log_4(sixteen) = 2 we get, log_4(x) -a million/2 log_4(x+3) =a million/2 so, 2 log_4(x) - log_4(x+3) =a million so, log _ 4 (x^2) - log _4(x+3) =a million so, log_4 ( x^2/(x+3)) = a million so, x^2/ (x+3) = 4 so, x^2 = 4x +12 so, x^2 -4x -12 =0 so, (x-6)(x+2)=0 so, x=6 (We could desire to discard the answer x=-2, via fact the log function is purely defined for helpful numbers.

log[4] (x) - log[4] (x - 1) = 1/2

log[4] (x/(x - 1)) = 1/2

x/(x - 1) = 4^(1/2)

x/(x - 1) = 2

x = 2(x - 1)

x = 2x - 2

x = 2

*LOL, nobody seems to realize that this is log base 4, XD

It is Sqrt(E) / (-4 + Sqrt(E)) = -0.701