Question 745948
log x + log (3x + 5)= 1
log x(3x + 5)= 1
x(3x + 5)= 10^1
x(3x + 5)= 10
3x^2 + 5x= 10
3x^2 + 5x - 10 = 0
Applying the "quadratic formula" yields:
x = {1.17, -2.84}
throw out the negative solution (extraneous) leaving:
x = 1.17
.
Details of "quadratic formula" follows:
*[invoke quadratic "x", 3, 5, -10 ]