Question 622941
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Hi,
these EQuations lend to the same method of solving:
Using Like bases with 'Equal' Exponents
Will do a couple of Exs:
 {{{125^(3x-4)=25^(4x+2)}}}
 {{{(5^3)^(3x-4) = (5^2)^(4x-2)}}}
 {{{ 5^(3(3x-4)) = 5^(2(4x+2))}}} 
       3(3x-4) = 2(4x+2)  |solve for x

{{{(1/64)^(0.5x-3) =8^(9x-2) }}}
{{{(8^(-2))^(0.5x-3) =8^(9x-2) }}}
{{{8^(-2(0.5x-3)) =8^(9x-2) }}}
    -2(.5x-3) = 9x-2     |solve for x