How to solve lg(3) ( 8x + 9) =  lg(9) (4x + 3)?  

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

Here we have to solve lg (3) (8x + 9) = lg (9) (4x + 3).

Now we use the property that lg (a) b = lg (n) b / lg (n) a, where n can be any positive number.

lg (3) (8x + 9) = lg (9) (4x + 3).

=> lg (3) (8x + 9) = lg (3) (4x + 3) / [lg (3) 9]

=> lg (3) (8x + 9) = lg (3) (4x + 3) / [lg (3) 3^2]

Now log a^b = b*log a

=> lg (3) (8x + 9) = lg (3) (4x + 3) / [2* lg (3) 3]

=> lg (3) (8x + 9) = lg (3) (4x + 3) / 2

=> 2* lg (3) (8x + 9) = lg (3) (4x + 3)

=> lg (3) (8x + 9) ^2 = lg (3) (4x + 3)

take the antilog of both the sides

=> (8x + 9) ^2 = (4x + 3)

=> 64x^2 + 81 + 144x – 4x – 3 = 0

=> 64x^2 + 140x + 78 = 0

=> 32x^2 + 70x + 39 = 0

But we see that the solutions of x are not real. And we do not have the logarithm for complex numbers.

Therefore the equation has no solutions.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll change the base of logarithm:

log (3) (4x + 3) = log (9) (4x + 3)*log (3) (9)

log (3) (4x + 3) = log (9) (4x + 3)*log (3) (3^2)

We'll apply the power rule of logarithms:

log (3) (3^2) = 2 log 3 3 = 2

log (9) (4x + 3) = [log (3) (4x + 3)]/2

We'll re-write the equation:

log (3) ( 8x + 9) = [log (3) (4x + 3)]/2

We'll cross multiply:

2log (3) ( 8x + 9) = [log (3) (4x + 3)]

log (3) ( 8x + 9)^2 = [log (3) (4x + 3)]

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

( 8x + 9)^2 = (4x + 3)

We'll expand the square from the left side:

64x^2 + 144x + 81 = 4x + 3

We'll subtract 4x + 3 both sides:

64x^2 + 144x + 81 - 4x  -  3 = 0

We'll combine like terms:

64x^2 + 140x + 78 = 0

32x^2 + 70x + 39 = 0

We'll apply the quadratic formula:

x1 = [-70 + sqrt(4900 - 4992)]/2*32

Since delta is negative; delta = -92, the equation has no real roots.

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