3=lnx
e^3=x
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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
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
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.
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).]
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