```Question 560460
{{{ log(3, (7x + 4)) - log(3,2) = 2*log(3,x) }}}
Use the rule
{{{ log(a) - log(b) = log(a/b) }}}
{{{ log(3,  ((7/2)*x + 2) ) = 2*log(3,x) }}}
Use the rule
{{{ a*log(b) = log(b^a) }}}
{{{ log(3, ((7/2)*x + 2) ) = log(3,x^2) }}}
{{{ (7/2)*x + 2 = x^2 }}}
{{{ x^2 - (7/2)*x - 2 = 0 }}}
{{{ 2x^2 - 7x - 4 = 0 }}}
{{{ x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = 2 }}}
{{{ b = -7 }}}
{{{ c = -4 }}}
{{{ x = (-(-7) +- sqrt( (-7)^2-4*2*(-4) ))/(2*2) }}}
{{{ x = ( 7 +- sqrt( 49 + 32 )) / 4 }}}
{{{ x = ( 7 +- sqrt( 81 )) / 4 }}}
{{{ x = ( 7 + 9 )/4 }}}
{{{ x = 4 }}}
and, taking the (-) square root,
{{{ x = ( 7 - 9 ) / 4 }}}
{{{ x = -2/4 }}}
{{{ x = -(1/2) }}}
--------------
Using {{{ x = 4 }}},
{{{ log(3, (7x + 4)) - log(3,2) = 2*log(3,x) }}}
{{{ log(3, (7*4 + 4)) - log(3,2) = 2*log(3,4) }}}
{{{ log(3, 32) b- log(3, 2) = 2*log(3,4) }}}
{{{ log(3, (32/2)) = log(3,4^2) }}}
{{{ log(3,16) = log(3,16) }}}
OK
Using {{{ x = -(1/2) }}}
{{{ log(3, (7*(-1/2) + 4)) - log(3,2) = 2*log(3,(-1/2)) }}}
I have to reject this solution for {{{x}}} because of the
right side. It is impossible to raise the base {{{3}}} to
some power ( + or - ) that will result in {{{ -(1/2) }}},
or any negative number.
So, {{{ x = 4 }}} is the only solution```