Question 577682
{{{ 9^(2x) * 9 = (27^(2x))/27 }}}
I would rewrite this as:
{{{ 9^( 2x + 1) = 27^(2x) * 27^(-1) }}}
{{{ 9^( 2x + 1 ) = 27^( 2x - 1 ) }}}
Now I can say {{{ 27 = 9^(3/2) }}}
This is true because {{{ 9^(1/2) = 3 }}} and
{{{ 3^3 = 27 }}}
{{{ 9^( 2x + 1 ) = ( 9^(3/2))^( 2x - 1 ) }}}
{{{ 9^( 2x + 1 ) = 9^( 3x - 3/2 ) }}}
Take the log to base 9 of both sides
{{{ 2x + 1 = 3x - 3/2 }}}
{{{ x = 2/2 + 3/2 }}}
{{{ x = 5/2 }}}
check answer:
{{{ 9^(2x) * 9 = (27^(2x))/27 }}}
{{{ 9^(2*5/2) * 9 = (27^(2*5/2))/27 }}}
{{{ 9^5 * 9 = 27^5 / 27 }}}
{{{ 9^6 = 27^4 }}}
{{{ 9^6 = (3^3)^4 }}}
{{{ (3^2)^6 = 3^12 }}}
{{{ 3^12 = 3^12 }}}
OK