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find the solutions of the following equations

1)  e2x= 8ex+20
2) log5(2x)+log5(x+1)=log5(3x)
3) log4(x+1)=2+log4(x-1)
please show step by step

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Michael F. | Mathematics TutorMathematics Tutor
4.7 4.7 (6 lesson ratings) (6)
1) e2x= 8ex+20 is a quadratic equation.  Let u=ex the equation is u2-8u-20=0 so  (u+2)(u-10)=0
The solutions are u=-2, ex=-2  not possible for a real value of x, u=10  ex=10 gives x=ln10

2) log5(2x)+log5(x+1)=log5(3x) so  log5((2x)(x+1))-log5(3x)=0  or log5((2x)(x+1)/3x)=0 or that
(2x)(x+1)/3x=50=1  gives 2x2+2x-3x=0  or 2x2-x=0 whose solutions are x(2x-1)=0 or x=0 x=1/2 But x=0 not possible
3) log4(x+1)=2+log4(x-1)  log4((x+1)/(x-1))=2  or (x+1)/(x-1)=42=16
x+1=16x-16  17=15x  x=17/15