Answer to Question #241644 in Algebra for Sachi

Question

Solve each of the following equations:

(i) e2x · (e2x-1)3 = 9

(ii) 2 ln x - ln(3 - 5x) = ln 2

Answer 1

(i)

$$e^{2x}\cdot(e^{2x}-1)^3=9$$

Use substitution $e^{2x}=t, t>0.$ Then

$$t(t-1)^3=9$$

$$t^4-3t^3+3t^2-t-9=0$$

$t>0=>t\approx2.527$

$$e^{2x}\approx2.527$$

$$x\approx0.5\ln2.527$$

$$x\approx0.464$$

(ii)

$$2\ln x-\ln (3-5x)=\ln 2$$

$x>0, 3-5x>0$

Then $0<x<\dfrac{3}{5}$

$$\ln x^2=\ln(2(3-5x))$$

$$x^2=6-10x$$

$$x^2+10x-6=0$$

$$D=(10)^2-4(1)(-6)=124$$

$$x=\dfrac{-10\pm\sqrt{124}}{2(1)}=-5\pm\sqrt{31}$$

$$-5+\sqrt{31}<\dfrac{3}{5}$$

$$\sqrt{31}<\dfrac{28}{5}$$

$$31<\dfrac{784}{25}$$

$$31<31.36, True$$

Answer: $x=\sqrt{31}-5.$