[tex]\bf 2log(x-3)+1=5\iff 2log_{10}(x-3)+1=5\\\\\\ 2log_{10}(x-3)=4
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log_{10}(x-3)=\cfrac{4}{2}\implies log_{10}(x-3)=2\\\\
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{{ a}}^{log_{{ a}}x}=x\impliedby \textit{using this cancellation rule}\\\\
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10^{\cfrac{}{}log_{10}(x-3)}=10^2\implies x-3=10^2\implies x=100+3[/tex]
