Question 197635
a)


{{{s(h)=2*ln(100h)}}} Start with the given equation.



{{{s(2h)=2*ln(100(2h))}}} Replace "h" with "2h" (since the altitude doubles)



{{{s(2h)=2*ln(2*100h)}}} Rearrange the terms.



{{{s(2h)=2(ln(2)+ln(100h))}}} Expand the log using the identity  {{{ln(A*B)=ln(A)+ln(B)}}}



{{{s(2h)=2*ln(2)+2*ln(100h)}}} Distribute



{{{s(2h)=2*ln(2)+s(h)}}} Replace {{{2*ln(100h)}}} with {{{s(h)}}}. Note: this can be done since {{{s(h)=2*ln(100h)}}}



{{{s(2h)=ln(2^2)+s(h)}}} Rewrite the log using the identity  {{{y*ln(x)=ln(x^y)}}}



{{{s(2h)=ln(4)+s(h)}}} Square 2 to get 4



So when the altitude doubles, the speed of the wind increases by {{{ln(4)}}} knots. Since *[Tex \LARGE \ln(4) \approx 1.38629], this means that the speed of the wind increases by about 1.38629 knots.



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b)


Are you only supposed to convert to a common log? If so, then there's not much to this part...



{{{s(h)=2*ln(100h)}}} Start with the given equation.



{{{s(h)=2*log(e,(100h))}}} Convert to a standard log of base "e"



{{{s(h)=2*(log(10,(100h))/log(10,(e)))}}} Now use the change of base formula (this step may be optional)



Remember the change of base formula is {{{log(b,(x))=log(10,(x))/log(10,(b))}}}