Question Number 208384 by efronzo1 last updated on 14/Jun/24
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Answered by A5T last updated on 14/Jun/24
$${log}_{{abc}} \left({a}\right)+{log}_{{abc}} \left({b}\right)=\mathrm{2}+\mathrm{3}=\mathrm{5}\Rightarrow{log}_{{abc}} \left({ab}\right)=\mathrm{5} \\ $$$${log}_{{abc}} \left({abc}\right)=\mathrm{1}={log}_{{abc}} \left({ab}\right)+{log}_{{abc}} \left({c}\right)=\mathrm{5}+? \\ $$$$\Rightarrow?=\mathrm{1}−\mathrm{5}=−\mathrm{4} \\ $$
Answered by Rasheed.Sindhi last updated on 14/Jun/24
$$\begin{cases}{\mathrm{log}_{{abc}} \left({a}\right)=\mathrm{2}\:}\\{\mathrm{log}_{{abc}} \left({b}\right)=\mathrm{3}\:}\\{\mathrm{log}_{{abc}} \left({c}\right)=?\:}\end{cases}\Rightarrow\begin{cases}{\left({abc}\right)^{\mathrm{2}} ={a}}\\{\left({abc}\right)^{\mathrm{3}} ={b}}\\{\left({abc}\right)^{{x}} ={c}}\end{cases} \\ $$$$\left({abc}\right)^{\mathrm{2}+\mathrm{3}+{x}} ={abc} \\ $$$$\mathrm{5}+{x}=\mathrm{1}\Rightarrow{x}=−\mathrm{4} \\ $$