(i)
e 2 x ⋅ ( e 2 x − 1 ) 3 = 9 e^{2x}\cdot(e^{2x}-1)^3=9 e 2 x ⋅ ( e 2 x − 1 ) 3 = 9 Use substitution e 2 x = t , t > 0. e^{2x}=t, t>0. e 2 x = t , t > 0. Then
t ( t − 1 ) 3 = 9 t(t-1)^3=9 t ( t − 1 ) 3 = 9
t 4 − 3 t 3 + 3 t 2 − t − 9 = 0 t^4-3t^3+3t^2-t-9=0 t 4 − 3 t 3 + 3 t 2 − t − 9 = 0
t > 0 = > t ≈ 2.527 t>0=>t\approx2.527 t > 0 => t ≈ 2.527
e 2 x ≈ 2.527 e^{2x}\approx2.527 e 2 x ≈ 2.527
x ≈ 0.5 ln 2.527 x\approx0.5\ln2.527 x ≈ 0.5 ln 2.527
x ≈ 0.464 x\approx0.464 x ≈ 0.464
(ii)
2 ln x − ln ( 3 − 5 x ) = ln 2 2\ln x-\ln (3-5x)=\ln 2 2 ln x − ln ( 3 − 5 x ) = ln 2 x > 0 , 3 − 5 x > 0 x>0, 3-5x>0 x > 0 , 3 − 5 x > 0
Then 0 < x < 3 5 0<x<\dfrac{3}{5} 0 < x < 5 3
ln x 2 = ln ( 2 ( 3 − 5 x ) ) \ln x^2=\ln(2(3-5x)) ln x 2 = ln ( 2 ( 3 − 5 x ))
x 2 = 6 − 10 x x^2=6-10x x 2 = 6 − 10 x
x 2 + 10 x − 6 = 0 x^2+10x-6=0 x 2 + 10 x − 6 = 0
D = ( 10 ) 2 − 4 ( 1 ) ( − 6 ) = 124 D=(10)^2-4(1)(-6)=124 D = ( 10 ) 2 − 4 ( 1 ) ( − 6 ) = 124
x = − 10 ± 124 2 ( 1 ) = − 5 ± 31 x=\dfrac{-10\pm\sqrt{124}}{2(1)}=-5\pm\sqrt{31} x = 2 ( 1 ) − 10 ± 124 = − 5 ± 31
− 5 + 31 < 3 5 -5+\sqrt{31}<\dfrac{3}{5} − 5 + 31 < 5 3
31 < 28 5 \sqrt{31}<\dfrac{28}{5} 31 < 5 28
31 < 784 25 31<\dfrac{784}{25} 31 < 25 784
31 < 31.36 , T r u e 31<31.36, True 31 < 31.36 , T r u e
Answer: x = 31 − 5. x=\sqrt{31}-5. x = 31 − 5.
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