Question 46644
{{{4^(2x)+1=2^(3x)+6}}}
{{{2xlog4+log1=3xlog2+log6}}}
{{{log1-log6=xlog2}}}
{{{log(1/6)=xlog2}}}
{{{log(1/6)=xlog(2)}}}
{{{x=(log(1/6))/(log(2))}}}
{{{x=(log(1/6))/(log(2))}}}
{{{x=2.585}}} to 3 decimal places.

I hope this helps,
Adam.
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