Question 733342
{{{ 32^( 4x-1 ) = 16^(3x) }}} is this it? If so,
Note that {{{ 32 = 2^5 }}} and {{{ 16 = 2^4 }}}
Now write this as:
{{{ ( 2^5 )^( 4x-1 ) = ( 2^4 )^( 3x ) }}}
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Now use the general rule:
{{{ (a^b)^c  = a^(b*c) }}}
{{{ 2^( 5*( 4x-1 ) ) = 2^( 4*3x ) }}}
For the equation to be true, the exponents on
both sides have to be equal, so
{{{ 5*( 4x-1 ) = 12x }}}
{{{ 20x - 5 = 12x }}}
{{{ 8x = 5 }}}
{{{ x = 5/8 }}}
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check:
{{{ 32^( 4x-1 ) = 16^(3x) }}}
{{{ 32^( 4*(5/8) - 1 ) = 16^(3*(5/8)) }}}
{{{ 32^( 5/2 - 1 ) = 16^( 15/8 ) }}}
{{{ 32^( 3/2 ) = 16^( 15/8 ) }}}
square both sides
{{{ 32^3 = 16^( 15/4 ) }}}
{{{ 32^3 = 2^15 }}}
{{{ 32768 = 32768 }}}
OK