Question 1139113
let m = mu.


.6 = e^(m * 2.1)


take the natural log of both sides of the equation to get:


ln(.6)) = ln(e^(m * 2.1))


since ln(e^x) = x * ln(e) and since ln(e) = 1, the equation becomes:


ln(.6) = m * 2.1


solve for m to get m = ln(.6) / 2.1 = -.243250297


replace m in the original equation with this value to see if the original equation holds true.


original equation is .6 = e^(m * 2.1) which becomes .6 = e^(-243250297 * 2.1) which becomes .6 = .6.


this confirms the solution is good.


the solution is that m = -.243250297.


here's a reference on log functions that might be helpful.


<a href = "https://www.purplemath.com/modules/solvelog.htm" target = "_blank">https://www.purplemath.com/modules/solvelog.htm</a>


here's another reference that might be helpful.


<a href = "http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/index.htm" target = "_blank">http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/index.htm</a>


check out tutorials 42 through 48.


any questions, write to dtheophilis@gmail.com