Question 824077
{{{ log( 7, x^2 ) = p }}}
I can say:
{{{ 7^p = x^2 }}}
{{{ x = ( 7^p )^(1/2) }}}
(1) {{{ x = 7^(p/2) }}}
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{{{ log( 7, x*y^2 ) = q }}}
I can say:
{{{ 7^q = x*y^2 }}}
{{{ y^2 = ( 7^q ) / x }}}
(2) {{{ y = ( 7^(q/2)) / x^(1/2) }}}
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From (1),
{{{ x^(1/2) = 7^(p/4) }}}
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(2) {{{ y = ( 7^(q/2)) / 7^(p/4) }}}
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{{{ x*y = ( 7^(p/2) )*( 7^(q/2) ) / 7^( p/4)  }}}
{{{ x*y =  ( 7^(2p/4) )*( 7^(q/2) ) / 7^( p/4)  }}}
{{{ x*y =  ( 7^( p/4) )*( 7^(q/2) ) }}}
{{{ x*y = 7^( p/4 + q/2 ) }}}
{{{ ( x*y )^(1/3) = 7^( p/12 + q/6 ) }}}
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Rewriting this:
{{{ log( 7, (x*y)^(1/3) ) = p/12 + q/6 }}}
Hope I got it