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.