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2log(x-3)+1=5

can someone explain how to solve for x?

User Livea
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\bf 2log(x-3)+1=5\iff 2log_(10)(x-3)+1=5\\\\\\ 2log_(10)(x-3)=4 \\\\\\ log_(10)(x-3)=\cfrac{4}{2}\implies log_(10)(x-3)=2\\\\ -----------------------------\\\\ {{ a}}^{log_{{ a}}x}=x\impliedby \textit{using this cancellation rule}\\\\ -----------------------------\\\\ 10^{\cfrac{}{}log_(10)(x-3)}=10^2\implies x-3=10^2\implies x=100+3
User Vincenzo Maggio
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