0
They both pass through the point (1,0) and have the same general shape. The log(x) curve is less steep than ln(x).
Wiki User
∙ 11y ago
Add your answer:
Earn +20 pts
What is the value of loge if base is not equal to e?
You can calculate log to any base by using: logb(x) = ln(x) / ln(b) [ln is natural log], so if you have logb(e) = ln(e) / ln(b) = 1 / ln(b)
What is the value of log -2?
[ln(2) + i*pi]/ln(10) if you are referring to log as a base 10 log. ln refers to thenatural logarithm (log base e)The log of any negative number is imaginary. The formula above is derived fromthe relationship:-1 = ei*pisince you want log of -2, multiply both sides by 2-2 = 2*ei*pitaking natural logarithm of both sides: ln( -2) = ln(2*ei*pi ) = ln(2) + ln(ei*pi )which reduces to ln(2) + i*piIf you want log10 then divide both sides by ln(10)So log10(-2) = ln(-2)/ln(10) = [[ln(2) + i*pi]/ln(10)x = log (-2) = log10(-2)10x = -2Think about the smallest possible number you can put in for x.10-∞ = ?10-∞ = 1/10∞10∞ = ∞1/∞ = ?1/∞ = 0It is impossible to ever get 0 or a negative number because you will never reach infinity.log(-2) is undefined
How do you differentiate 2 to the power x with respect to x?
The answer is ln(2)2x where ln(2) is the natural log of 2. The answer is NOT f(x) = x times 2 to the power(x-1). That rule applies only when the exponent is a constant.
What is the derivative of y equals xlnx?
Use the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln x
If y equals log x what is dy over dx?
Assuming that is the natural logarithm (logarithm to base e), the derivative of ln x is 1/x. For other bases, the derivative of logax = 1 / (x ln a), where ln a is the natural logarithm of a. Natural logarithms are based on the number e, which is approximately 2.718.
Graph Inverse function of the exponential function?
An exponential function is of the form y = a^x, where a is a constant. The inverse of this is x = a^y --> y = ln(x)/ln(a), where ln() means the natural log.
How do you graph x to the power of x with both the real and non-real parts?
This depends on what 'space' you want to plot it on. Another thing, are the x values all complex, or just real. I did a writeup on a similar question awhile back only considering real values of x, but complex values of y. For the positive sided, one way is to take the log of x^x, then you have log(y) = x*log(x). You can plot this manually on log paper. If you use natural log, instead, then you have ln(y) = x*ln(x), then take e raised to both sides: y = e^(x*ln(x)). To take natural log of a negative number, consider this: if u = A*e^(iΘ), then ln(u) = ln(A) + (iΘ)*ln(e) = ln(A) + (iΘ)
Give an example of an exponential function Find this exponential function's inverse which will be a logarithmic function Plot the graph of both the functions?
An exponential function can be is of the form f(x) = a*(b^x). Some examples are f1(x) = 3*(10^x), or f2(x) = e^(-2*x). Note that the latter still fits the format, with b = e^(-2). The inverse is the logarithmic function. So for y = f1(x) = 3*(10^x), reverse the x & y, and solve for y:x = 3*(10^y)log(x) = log(3*(10^y)) = log(3) + log(10^y) = log(3) + y*log(10) = y*1 + log(3)y = log(x) - log(3) = log(x/3)The second function: y = e^(-2*x), the inverse is: x = e^(-2*y).ln(x) = ln(e^(-2*y)) = -2*y*ln(e) = -2*y*1y = -ln(x)/2 = ln(x^(-1/2))See related link for an example graph.
If e to the power x equals 0.4634 find x?
the natural log, ln, is the inverse of the exponential. so you can take the natural log of both sides of the equation and you get... ln(e^(x))=ln(.4634) ln(e^(x))=x because ln and e are inverses so we are left with x = ln(.4634) x = -0.769165
What is the value of loge if base is not equal to e?
You can calculate log to any base by using: logb(x) = ln(x) / ln(b) [ln is natural log], so if you have logb(e) = ln(e) / ln(b) = 1 / ln(b)
How do you solve log2 equals x?
log(2) = X can be expressed exponentially like this, because by the definition of logs( base 10) this is what this means. 10^X = 2 take natural log each side ln(10^X) = ln(2) you have right to place X in front of ln X ln(10) = ln(2) X = ln(2)/ln(10) ( not ln(2/10)!! ) X = 0.3010299957 check 10^0.3010299957 = 2 checks
Solve ln y equals xln e?
ln is the natural logarithm. That is it is defined as log base e. As we all know from school, log base 10 of 10 = 1 just as log base 3 of 3 = 1, so, likewise, log base e of e = 1 and 1.x = x. so we have ln y = x. Relace ln with log base e, and you should get y = ex
X minus 1 plus 2 plus log x equals 3. find x?
Original Statement:x - 1 + 2 + log(x) = 3Simplify:x + 1 + log(x) = 3Subtract 1:x + log(x) = 2Lambert W-Function:x = (W(100*ln(10))/(ln(10)) = 1.7555794993... (rounded up).This considered log(x) to be base 10 log (x).
How do I solve the equation e to the power of x equals 2 using logarithms?
When the logarithm is taken of any number to a power the result is that power times the log of the number; so taking logs of both sides gives: e^x = 2 → log(e^x) = log 2 → x log e = log 2 Dividing both sides by log e gives: x = (log 2)/(log e) The value of the logarithm of the base when taken to that base is 1. The logarithms can be taken to any base you like, however, if the base is e (natural logs, written as ln), then ln e = 1 which gives x = (ln 2)/1 = ln 2 This is in fact the definition of a logarithm: the logarithm to a specific base of a number is the power of the base which equals that number. In this case ln 2 is the number x such that e^x = 2. ---------------------------------------------------- This also means that you can calculate logs to any base if you can find logs to a specific base: log (b^x) = y → x log b = log y → x = (log y)/(log b) In other words, the log of a number to a given base, is the log of that number using any [second] base you like divided by the log of the base to the same [second] base. eg log₂ 8 = ln 8 / ln 2 = 2.7094... / 0.6931... = 3 since log₂ 8 = 3 it means 2³ = 8 (which is true).
Solve e raised to the power of negative x equals 6?
e-x = 6Take the natural log of both sides:ln(e-x) = ln(6)-x = ln(6)x = -ln(6)So x = -ln(6), which is about -1.792.
Differentiate log x?
The derivative of ln x, the natural logarithm, is 1/x.Otherwise, given the identity logbx = log(x)/log(b), we know that the derivative of logbx = 1/(x*log b).ProofThe derivative of ln x follows quickly once we know that the derivative of ex is itself. Let y = ln x (we're interested in knowing dy/dx)Then ey = xDifferentiate both sides to get ey dy/dx = 1Substitute ey = x to get x dy/dx = 1, or dy/dx = 1/x.Differentiation of log (base 10) xlog (base 10) x= log (base e) x * log (base 10) ed/dx [ log (base 10) x ]= d/dx [ log (base e) x * log (base 10) e ]= [log(base 10) e] / x= 1 / x ln(10)
128 equals x to the power 7 what is x how do i use logarithms to solve it?
128 = x7 (we see that x > 0, since x is raised to an odd power) 27 = x7 (this is true only when x = 2) 2 = x Remember that a logarithm is an exponent. The statement 128 = x7 is equivalent to logx 128 = 7. logx 128 = 7 log 128/log x = 7 (or reverse the both sides) log x/log 128 = 1/7 (multiply by log 128 to both sides) log x = log128/7 (or use base 10 for the logarithm) log10 x = log128/7 (write the equivalent statement) 10log 128/7 = x 2 = x or use the natural log, ln. 128 = x7 ln 128 = ln x7 ln 128 = 7ln x (ln 128)/7 = ln x e(ln128)/7 = elnx (elnx = x ) 2 = x
