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Solve for x:  lg(8x+9) + lgx = 1 + lg(x^2 - 1)

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merishorverde | Student, College Freshman | eNoter

Posted January 3, 2011 at 11:30 PM via web

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Solve for x:

 lg(8x+9) + lgx = 1 + lg(x^2 - 1)

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giorgiana1976 | College Teacher | Valedictorian

Posted January 3, 2011 at 11:37 PM (Answer #1)

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We'll impose the constraints of existence of logarithms:

8x+9 > 0

x > -9/8

x^2 - 1 > 0

(x-1)(x+1) > 0

x > 1

x > -1

The interval of admissible values for x is (1 ; +infinite).

Now, we'll solve the equation:

lg (8x+9) + lgx = 1 + lg(x^2 - 1)

We'll write 1 = lg 10 and we'll apply the product rule of logarithms:

lg [x*(8x+9)] = lg 10(x^2 - 1)

Since the bases are matching, we'll apply one to one property:

 [x*(8x+9)] = 10(x^2 - 1)

We'll remove the brackets:

8x^2 + 9x = 10x^2 - 10

We'll subtract 8x^2 + 9x and we'll use symmetric property:

2x^2 - 9x - 10 = 0

We'll apply quadratic formula:

x1 = [9 + sqrt(81 + 80)]/4

x1 = (9+sqrt161)/4

x2 = (9-sqrt161)/4

Since the value of x2 doesn't belong to the interval of admissible values, we'll reject x2.

The only solution of the equation is x1 = (9+sqrt161)/4.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted January 3, 2011 at 11:46 PM (Answer #2)

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We have to solve lg(8x+9) + lgx = 1 + lg(x^2 - 1) for x.

Now, we use the relation that lg a + lg b = lg(a*b)

lg(8x+9) + lgx = 1 + lg(x^2 - 1)

=> lg [( 8x + 9)*x] = 1 + lg ( x^2 - 1)

=> lg [ 8x^2 + 9x] - lg ( x^2 - 1) = 1

=> lg[ ( 8x^2 + 9x) / (x^2 - 1)] = 1

If we take the base of the log to be 10

=> ( 8x^2 + 9x) / (x^2 - 1) = 10

=> ( 8x^2 + 9x) = 10x^2 - 10

=> 2x^2 - 9x - 10 =0

x1 = [-b + sqrt (b^2 – 4ac)]/2a

=> [ 9 + sqrt ( 81 + 80)/4]

=> 9/4 + sqrt (161)/4

x2 = 9/4 - sqrt 161 / 4, this is ignored as it is negative.

The value of x is 9/4 + sqrt 161 /4.

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted January 4, 2011 at 12:09 AM (Answer #3)

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Given the logarithm equation:

lg (8x+9) + lg (x) = 1+ lg (x^2-1)

We need to find x value.

We will use logarithm properties to solve.

First, we know that lg a + lg b = lg (ab)

==> lg (x(8x+9) = 1 + lg (x^2 -1)

Also, we know that lg 10 = 1

==> lg (8x^2 + 9x) = lg 10 + lg (x^2-1)

==> lg (8x^2 +9x) = lg 10(x^2-1)

==> lg (8x^2 +9x) = lg (10x^2 -10)

Now we have the logs are equal, then the bases are equal too.

==> 8x^2 + 9x = 10x^2 - 10

We will combine like terms.

==> 2x^2 - 9x -10 = 0

Now we will find the roots.

==> x1= ( 9 + sqrt(81+80) / 4 = (9+sqrt(161) / 4

==> x2= (9-sqrt(161) / 4 ( Not valid)

Then, the answer is:

x = (9 + sqrt161) /4

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neela | High School Teacher | Valedictorian

Posted January 4, 2011 at 12:58 AM (Answer #4)

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Solve for x: lg(8x+9) + lgx = 1 + lg (x^2 -1).

lga+lg b = lg ab. 1= log10

Therefore we can rewrite the equation as below:

lg(8x+9)x = lg10+lg(x^1-1)

lg((8x^2+9x) = lg(10x^2-10).

Wetake antilog:

8x^2+9x = 10x^2-10.

0 = 10x^2-10 - 8x^2-9x.

0 = 2x^2 -9x -10 .

2x^2-9x-10 = 0.

 x1 = {9 + sqrt(9^2-4*2*-10)}/2*2 = (9+sqrt161)/4

x2 = {9- sqrt161}/4.

Therefore the solutions of the equation are x1 = (9+sqrt161)/4 and x2 = (9-sqrt161)/4.

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