SOLUTION: solve for x: log base 3 (7x+4)-log base 3 of 2 = 2logbase3 of X

Question 560460: solve for x:
log base 3 (7x+4)-log base 3 of 2 = 2logbase3 of X
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
solve for x:
log base 3 (7x+4)-log base 3 of 2 = 2logbase3 of X
----
log3(7x+4) - log3(2) = log3(x^2)
-------
log3[(7x+4)/2] = log3(x^2)
------
[(7x+4)/2] = x^2
----
2x^2 - 7x - 2 = 0
---
x = [7 +- sqrt(49-4*2*-2)]/4
-------
x = [7 +- sqrt(49+16)]/4
----
Positive solution:
-----
x = [7 + sqrt(65)]/4
----
x = 3.7656..
================
Cheers,
Stan H.

Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!

Use the rule

Use the rule

Use the quadratic formula

and, taking the (-) square root,

--------------
Using ,

OK
Using

I have to reject this solution for because of the
right side. It is impossible to raise the base to
some power ( + or - ) that will result in ,
or any negative number.
So, is the only solution
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