Question 1088005
Solve the following equation
A) 9 log 5 base x = log x base 5 
B) log x/2 base 8 = log x base 2

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A)  I don't see how to solve this without using an iterative technique
     {{{ 9log(x,(5)) = log(5,(x)) }}}
     {{{ log(x,(5)) = (1/9)*log(5,(x)) }}} 
   {{{ matrix(3,3,"","","","","","", x^(log(x,(5))), " = ", x^((1/9)*log(5,(x)) )) )  }}}
    {{{ matrix(3,3,"","","","","","", 5, " = ", x^((1/9)*log(5,(x)) )) )  }}}
    {{{ matrix(3,3,"","","","","","",   x^((1/9)*log(5,(x)) ) - 5, " = ", 0 )  }}}

—  Maybe another tutor has a better solution.  I would be tempted to iterate the last equation until it is (close enough to) zero.   

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This one is easily solved

B) {{{ log(8, (x/2)) = log(2,(x)) }}}
    {{{ matrix(2,3,"","","", 8^log(8,(x/2)), "=", 8^log(2,(x))) }}}
   {{{ matrix(2,3,"","","",  x/2 , "=", (2^3)^log(2,(x))) }}}
   {{{ matrix(2,3,"","","",  x/2 , "=", (2^log(2,(x)))^3 ) }}}
   {{{ matrix(1,3,  x/2 , "=", x^3 ) }}}
   {{{ matrix(1,3,  1/2 , "=", x^2 ) }}}
   {{{ highlight( matrix(1,3,  x , "=",  sqrt(2)/2) ) }}}

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7/18/17 - I can't believe I overlooked that property, {{{ log(a,b) = 1/(log(b,a)) }}}, which is easy enough to derive from the change-of-base calculation.    Thanks tutor Ikleyn!