Solve this equation for x log x 1 log x 1 1 The atmosphe

Solve this equation for x log (x + 1) + log (x - 1) = 1 The atmospheric pressure P, in pounds per square inch, at x miles above sea level is given by P(x) = 14.7e^-21x At what height will the atmospheric pressure be half of the sea level pressure? If the energy release of one earthquake is 1,000,000 times that of another, how much larger is the Richter scale reading of the larger than the smaller? That is, find M_2-M_1 Let M_1 = 2/3 log E_1/E_0 and M_2 = 2/3 log E_2/E_0 where E_2 = 1,000,000 E_1

Solution

8)log(x+1)+log(x-1) =1

log((x+1)(x-1)) =1

log(x²-1) =1

(x²-1) =10

x² =11

x =11

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9)given P(x)=14.7e^-0.21x

at sea level x =0

P(0)=14.7e^-0.21*0

P(0)=14.7

half of sea level pressure =14.7/2

14.7/2 =14.7e^-0.21x

e^-0.21x=(1/2)

e^0.21x=2

0.21x=ln2

x=(100_/21)ln2

x=3.3

at height of 3.3 miles atmospheric pressure is half that of at sea level

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10)M₂-M₁=(2_/3)log(E₂/E₀) -(2_/3)log(E₁/E₀)

M₂-M₁=(2_/3)[log(E₂/E₀) -log(E₁/E₀)]

M₂-M₁=(2_/3)[log((E₂/E₀)/(E₁/E₀))]

M₂-M₁=(2_/3)[log(E₂/E₁)]

M₂-M₁=(2_/3)[log(1000000E₁/E₁)]

M₂-M₁=(2_/3)[log1000000]

M₂-M₁=(2_/3)(6)

M₂-M₁=4