lg(x-1) + lg(6x-5) =2

We know that lgx + lgy = lg(xy)

==> lg(x-1) + lg(6x-5) = 2

==> lg(x-1)(6x-5) = 2

==> lg(6x^2-11x +5)= lg100

==> 6x^2 -11x +5 = 100

==> 6x^2 -11x -95=0

Factorize:

==> (x-5)(6x+19)=0

==> x1= 5

and x2= -19/6

To check answers:

substitute with x1=5

log(x-1) + log(6x-5)=2

log(5-1) + log(30-5)=2

log(4) + log(25) =2

log(4*25) =2

log(100)=2

log(10)^2=2

2log(10 =2

2*1 =2

2=2

Now substitute with x2= -19/6

log(-19/6 -1) + log(6(-19/6) -5) =2

The log has negative values , then we will nit consider the solution x2=-19/6 .

**Then the only solution is x1=5**

First we'll establish the constraints for existence of logarithms:

x-1>0

We'll add 1 both sides:

x>1

And

6x-5>0

We'll add 5 both sides:

6x>5

We'll divide by 6:

x>5/6

The value for x which satisfies both constraints is x>1.

Now, we'll solve the equation to find out the values for x.

We'll apply the product rule:

lg[(x-1)(6x-5)] = 2

We know that lg has the base 10, so we'll re-write the equation:

[(x-1)(6x-5)] = 10^2

We'll open the brackets:

6x^2 - 5x - 6x + 5 = 100

6x^2 - 11x + 5 - 100 = 0

We'll apply the quadratic formula:

x1 = [11+sqrt(121+24*95)]/12

x1 = (11+49)/12

x1 = 60/12

x1 = 5

x2 = (11-49)/12

x2 = -38/12

x2 = -19/6 < 1

Because from both solutions, only the first is bigger than 1, the equation will have only one solution, namely x = 5.

lg(x-1)+lg(6x-5) = 2. To find x.

Solution:

Since lga+lgb = lgab,

lg(x-1)+lg(6x-5) = lg((x-1)(6x-5)) = 2.

Taking antilog,

(x-1)(6x-5) = 10^2.

6x^2 -11x+5 = 100.

6x^2-11x+5-100 = 0.

6x^2-11x-95 = 0.

6x^2-30x+19x-95 = 0

6x(x-5)+19(x-5) = 0

(x-5)(6x+19) = 0.

x-5= 0 or 6x+19 = 0.

x=5 or x=-19/6