# Doubt about an equationI have doubts about the following equation: 2*lg x/lg(5x-4)=1 This has solutions x1 = 4 and x2 = 1,also. Why only x1 = 4 is the final solution?

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### 3 Answers

2*lg x/lg(5x-4)=1

Using the power property of log yields: 2lgx = lg(x^2)

lg(x^2)/lg(5x-4)=1 => lg(x^2)/lg(5x-4)-1 = 0

You need to bring the terms to a common denominator such that:

lg(x^2) - lg(5x-4)= 0

Using the quotient property of log yields:

lg ((x^2)/(5x-4)) = 0 => ((x^2)/(5x-4)) = 10^0 => ((x^2)/(5x-4)) = 1

Subtracting 1 both sides yields:

((x^2)/(5x-4)) - 1 = 0

You need to bring the terms to a common denominator such that:

x^2 - 5x + 4 = 0 => (x^2 - 4x) - (x - 4) = 0

x(x - 4) - (x - 4) = 0 => (x-4)(x-1)=0

If (x-4)(x-1)=0 => x - 4 = 0 => x = 4

x - 1 = 0 => x = 1

You need to plug x = 1 in the equation to verify if it may be a solution for it.

2*lg 1/lg(5-4)=1 => 2*lg1/lg1 = 1 => 2=1 contradiction

You need to plug x = 4 in the equation to verify if it may be a solution for it.

2*lg 4/lg(20-4)=1 => lg(4^2)/lg16 = 1 => lg 16/lg 16 = 1 => 1 = 1

**Hence, the solution to this equation is x = 4.**

It's obvious that we have to deal with a logarithmic equation, where conditions of existence are a"must", before start solving it!

So, conditions of existence are:

x>0

(5x-4)>0, 5x>4, x>4/5

From both inequation, the common interval of values which are satisfying them is: (4/5, infinity). So, it is obvious that the solution x2 = 1 doesn't belong to the interval of allowed values, (4/5, infinity).

2*lg x/lg(5x-4)=1

2*lg x=lg(5x-4)

lg (x^2)=lg(5x-4)

Because of the fact that the logarithmic function is injective, that means that (x^2)=(5x-4)

x^2-5x+4=0

x1 +x2=-(-5/1), x1+x2=5

x1*x2=4/1

So 1+4=5 and 1*4=4

x1=4, x2=1, but **x2=1doesn't belong to the set interval of allowed value, for the equation is possible, which is **

**(4/5, infinity). **

The equation, 2*lgx/lg(5x-4)=1 or

2lgx =lg(5x-4), on taking anti logarithms, becomes as below:

x^2=5x-4. This could be rewritten like:

x^2-5x+4=0 or , after factorisation,

(x-4)(x-1) = 0.

So x = 4 checks well with the function giving 2log4/(log(24-4) =log(16)/log(16) = 1.

But at x= 1 does not check well giving us 2log(1)/log(5*1-4) = log(1)log(1) = 0/0 form and the limiting value of the 2logx/log(5x-4) as x--> 1, by LHospital's rule is (2logx)'/[log(5x-4)]' as x-->1 = (10x-8)/(5x) as x-->1 = 0.4 and it is not 1. So x=1 does not satisfy the equation.