Question 1106348
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NOTE: you can write log base b of x as "log(b,x)".  Or perhaps better, enclose the x in parentheses, as "log(b,(x))"<br>
{{{log(2,(x-4))+log(2,(x))=5}}}
{{{log(2,(x-4))+log(2,(x)) = log(2,(32))}}}  [rewrite the constant 5 as log base 2 of some number, so that all the terms in the equation are log base 2 of something]
{{{log(2,((x-4)(x))) = log(2,(32))}}}  [sum of logs = log of the product]
{{{log(2,(x^2-4x)) = log(2,(32))}}}  [algebra...]
{{{x^2-4x = 32}}}  [if the logs of the expressions are equal, the expressions are equal]
{{{x^2-4x-32 = 0}}}
{{{(x-8)(x+4) = 0}}}
{{{x=8}}} or {{{x=-4}}}<br>
But x=-4 does not satisfy the original equation (log of a negative number is undefined).
x=8 satisfies the original equation, so the unique solution to the equation is x=8.