1-log5x=log5(x-1)

Question

Solve for x

Answer 1

Not sure how to read this.

If it's log to base 5 of x and of (x-1), the left hand side is 1- log[5](x) and the right hand side is log[5](x-1). So, since 1 is log[5](5), log[5](5/x)=log[5](x-1). Remove logs: 5/x=x-1. Multiply through by x: 5=x^2-x. Completing the square: x^2-x+1/4=5+1/4, (x-1/2)^2=21/4, x-1/2=sqrt(21)/2 and x=(1/2)(1+sqrt(21)) or (1/2)(1-sqrt(21)). However, the second solution makes x negative, and we can't take logs of a negative number. Therefore the solution is x=1/2(1+sqrt(21))=2.7913.

If the question is 1-ln(5x)=ln(5(x-1)), then ln(e/5x)=ln(5(x-1)), e/5x=5x-5, e=25x^2-25x, 25(x^2-x+1/4)=e+25/4, 5(x-1/2)=sqrt(e+6.25), x-1/2=sqrt(e+6.25)/5, x=1/2+(sqrt(e+6.25))/5=1.0989. For any other base replace e with the base.

Answer 2

1 - log5x   =    log5(x-1)

   +log5x       =log5(x - 1)

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1 = log5x + log5(x - 1)

1 = log5 (x(x-1))

log5(x^2 - x ) = 1

x^2 - x   =   5^1

        -5       - 5

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x^2 - x - 5 = 0

A = 1,   B = - 1 , C = - 5

  x = [  - b +(-) sgrt(b^2-4AC)]/2A

x = {-(-1)=(-)sgrt((-1)^2-4(-5))}/2

x =[1 +(-)sgrt21]/2

x = ( 1 + 4.58)/2          ,       x = ( 1 - 4.58)/2

 x = 2.79          and                          x = - 1.79 ( t,here is not a log of a negative number)

1 = log5((2.79*(2.78-1))              

1 = log5(5)

log5(5) = 1

1 = 1    : x = 2.79 is a solution

x = 2.79 is a solution