2x-2/x^2+3x-4
y = x / (x^2 + 2x + 1) The horizontal asymptote is y = 0
-1
[log2 (x - 3)](log2 5) = 2log2 10 log2 (x - 3) = 2log2 10/log2 5 log2 (x - 3) = 2(log 10/log 2)/(log5/log 2) log2 (x - 3) = 2(log 10/log 5) log2 (x - 3) = 2(1/log 5) log2 (x - 3) = 2/log 5 x - 3 = 22/log x = 3 + 22/log 5
log5x
Yes, the asymptote is x = 0. In order for logarithmic equation to have an asymptote, the value inside log must be 0. Then, 5x = 0 → x = 0.
2x-2/x^2+3x-4
7x = 5x log(7) = log(5)x = log(5) / (log(7) = 0.82709 (rounded)
The only way I ever learned to find it was to think about it. The function f(x) = log(x) only exists of 'x' is positive. As 'x' gets smaller and smaller, the function asymptotically approaches the y-axis.
y = x / (x^2 + 2x + 1) The horizontal asymptote is y = 0
It is y = 0
The explanation and answer to the following math equation to find x -0.3 plus 5-5 log (d) equals a plus 5-5 log 4 (d) is -5 log(d)+x+4.7 = a-5 log(4 d)+5. The solution is x = a-6.63147.
-1
Yes. Take the functions f(x) = log(x) or g(x) = ln(x) In both cases, there is a vertical asymptote where x = 0. Because a number cannot be taken to any power so that it equals zero, and can only come closer and closer to zero without actually reaching it, there is an asymptote where it would equal zero. Note that transformations (especially shifting the function left and right) can change the properties of this asymptote.
[log2 (x - 3)](log2 5) = 2log2 10 log2 (x - 3) = 2log2 10/log2 5 log2 (x - 3) = 2(log 10/log 2)/(log5/log 2) log2 (x - 3) = 2(log 10/log 5) log2 (x - 3) = 2(1/log 5) log2 (x - 3) = 2/log 5 x - 3 = 22/log x = 3 + 22/log 5
log5x
with something called logarithms. So 1 = (1 + x)^5 log 1 = log ((1+x)^5) log 1 = 5 x log (1 +x) but log 1 = 0 therefore 0 = 5 x log(1+x) divide both sides by 5 and you get 0 = log (1+x) we know that log 1 = 0, therefore 1+ x = 1 and so x = 0