Question 1198055

Solve the indicial linear equation 5√8^2x=128^x X 64
<pre>    {{{matrix(1,3, 5sqrt(8^(2x)), "=", 128^x * 64)}}}
{{{matrix(1,3, 5sqrt((2^3)^(2x)), "=", (2^7)^x * 64)}}} ------ Changing 8 and 128 to base 2
    {{{matrix(1,3, 5sqrt(2^(6x)), "=", matrix(2,1, " ", 2^(7x) * 64))}}} ------- Applying {{{matrix(1,3, (a^b)^c, "=", a^(bc))}}}
{{{matrix(1,3, 5(2^(6x))^(1/2), "=", matrix(2,1, " ", 2^(7x) * 64)))}}} ------- Applying {{{matrix(1,3, sqrt(a), "=", matrix(2,1, " ", a^(1/2))))}}}
   {{{matrix(1,3, 5(2^(3x)), "=", 2^(7x) * 64)}}} ------- Applying {{{matrix(1,3, (a^b)^c, "=", a^(bc))}}}
         {{{matrix(1,3, 5/64, "=", 2^(7x)/2^(3x))}}} 
         {{{matrix(1,3, 5/64, "=", 2^(4x))}}} --- Applying {{{matrix(1,3, a^b/a^c, "=", a^(b - c))}}}
         {{{matrix(1,3, 4x, "=", log(2, (5/64)))}}} ----- Converting to LOGARITHMIC form
         {{{matrix(1,3, 4x, "=", log((5/64))/log((2)))}}} ====> {{{matrix(1,3, x, "=", (log((5/64))/log((2)))/4)}}} =====> {{{matrix(1,5, highlight(x), "=", log((5/64))/4log((2)), "=", highlight(- 0.919517976))}}}</pre>