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**

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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