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Q: Does 5 equals log 5 to the x have an asymptote?
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Does y equals log5x have an asymptote?

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.


Which function has the following a vertical asymptote at x equals -4 horizontal asymptote at y equals 0 and a removable discontinuity at x equals 1?

2x-2/x^2+3x-4


How would you rewrite 7 to the power of x equals 5?

7x = 5x log(7) = log(5)x = log(5) / (log(7) = 0.82709 (rounded)


How do you find the vertical asymptote for the logarithmic function f of x equals log x?

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.


What is the horizontal asymptote of y equals x divided by x2 plus 2x plus 1?

y = x / (x^2 + 2x + 1) The horizontal asymptote is y = 0


What is the horizontal asymptote of y equals 2 to the power of x?

It is y = 0


Can anyone give an explanation and an answer to the following Find x -0.3 plus 5-5logd equals a plus 5-5log4d?

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.


What is the horizontal asymptote of 5 divided by the quantity of x minus 6?

-1


Do logarithmic functions have vertical asymptotes?

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.


How do you solve log base 2 of x - 3 log base 2 of 5 equals 2 log base 2 of 10?

[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


What is the inverse of the function y equals log 5 x?

log5x


How do you isolate x in this equation 1 equals leftbracket 1 plus x right bracket exponent 5?

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