Question 1106333
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Note you can write "log base 2 of x" as "log(2,x)", or, better yet, "log(2,(x))".<br>
{{{log(2,(6x-1))-3 = 2log(2,(x))}}}
{{{log(2,(6x-1))-log(2,(8)) = 2log(2,(x))}}}  [write the constant 3 as log base 2 of something]
{{{log(2,((6x-1)/8)) = log(2,(x^2))}}}  [rules of logarithms: difference of logs = log of quotient; log(x^n) = n*log(x)]
{{{(6x-1)/8 = x^2}}}  [if the logs of the expressions are equal, then the expressions are equal]
{{{6x-1 = 8x^2}}}
{{{8x^2-6x+1 = 0}}}
{{{(4x-1)(2x-1) = 0}}}
{{{x = 1/4}}} or {{{ x = 1/2}}}<br>
Both solutions satisfy the original equation, so both are solutions.