Solve the indicia equation 3(2^2x+3)-5(2^x+2=0

the question is a technical one I hope it can be solved step by step

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1 Answer

Let y=2^x: 3(y^2+3)-5(y+2)=0.

3y^2+9-5y-10=0.

3y^2-5y-1=0; y=(5±√(25+12))/6.

We can only take the positive root: 1.8471 approx.

So 2^x=1.8471;

taking logs: xlog2=log1.8471, x=log1.8471/log2=0.8853 approx.

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