How do you solve $log_2x + log_2(x+1) = log26$ ?

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Answer 1 · 2015-10-30T11:31:45

$x=1,2078$

By using the laws of logs, we may rewrite this equation as

$log_2[x*(x+1)]=log_(10)26$

$thereforelog_2(x^2+x)=log_(10)26$

$thereforex^2+x=2^(log26)=2,6665481$

This is now a quadratic equation which we may use the quadratic formula to solve for:

$thereforex=(-1+-sqrt(1-(4xx1xx-2,6665481)))/2$

$=1,2078 or -2,2078$

But $x=-2,2078!inDomlog_2x$ and so $x=1,2078$ is the only feasible solution.

How do you solve log_2x + log_2(x+1) = log26log_2x + log_2(x+1) = log26?