find the solutions of the following equations

Question

1)  e^2x= 8e^x+20

2) log5(2x)+log5(x+1)=log5(3x)

3) log4(x+1)=2+log4(x-1)

 

please show step by step

Answer 1

1) e^2x= 8e^x+20 is a quadratic equation.  Let u=e^x the equation is u²-8u-20=0 so  (u+2)(u-10)=0

The solutions are u=-2, e^x=-2  not possible for a real value of x, u=10  e^x=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