log3(8x+3) = 1+ log3 (x^2)
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We have : log3(8x+3) = 1+ log3 (x^2)
log3(8x+3) = 1+ log3 (x^2)
=> log3(8x+3) - log3 (x^2) = 1
use the property log a - log b = log (a/b)
=> log3 [(8x + 3)/(x^2)] = 1
=> (8x + 3)/(x^2) = 3
=> 8x + 3 = 3x^3
=> 3x^2 - 8x - 3 = 0
=> 3x^2 - 9x + x - 3 = 0
=> 3x(x - 3) + 1(x - 3) = 0
=> (3x + 1)(x - 3)= 0
=> x = -1/3 and x = 3
Both the values are defined for the logs given.
The solution of the equation is x = -1/3 and x = 3
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The equation `log_3(8x+3) = 1+ log_3 (x^2)` has to be solved for x.
`log_3(8x+3) = 1+ log_3 (x^2)`
=> `log_3(8x+3) - log_3 (x^2) = 1`
Use the property log a - log b = log(a/b)
=> `log_3((8x+3)/(x^2)) = 1`
Use the property `log_b b = 1`
This gives: `(8x+3)/(x^2) = 3`
8x + 3 = 3x^2
3x^2 - 8x - 3 = 0
3x^2 - 9x + x - 3 = 0
3x(x - 3) + 1(x - 3) = 0
(3x + 1)(x - 3) = 0
3x + 1 = 0, x = -1/3
and x - 3 = 0, x = 3
Both the values x = -1/3 and x = 3 satisfy the given equation
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