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What is the equation for exponential interpolation?
The correct formula for exponential interpolation is: y =ya*(yb/ya)^[(x-xa)/(xb-xa)], xa<x<xb and also, x=xa*[ln(yb)-ln(y)]/[ln(yb)-ln(ya)]+xb*[ln(y)-ln(ya)]/[ln(yb)-ln(ya)], ya<y<yb
How would you solve ln 4 plus 3 ln x equals 5 ln 2?
Ln 4 + 3Ln x = 5Ln 2 Ln 4 + Ln x3= Ln 25 = Ln 32 Ln x3= Ln 32 - Ln 4 = Ln (32/4) = Ln 8= Ln 2
Is there a function where the first derivative goes to infinity for x going to 0 and where the first derivative equals 0 when x is 1?
Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.
What is the logarithmic form of the equation e4x 2981?
Assuming your equation is: e^4x = 2981 Then taking the logs to base e (natural logs) of both sides gives: ln(e^4x) = ln(2981) → 4x = ln(2981) [Note: The 2981 looks like it has been rounded to a whole number from approx 2980.958 which means the value of x was a whole number (to which ln(2981) rounds).]
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
