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Question-191155




Question Number 191155 by mathlove last updated on 19/Apr/23
Answered by mahdipoor last updated on 19/Apr/23
log(xy)=log(x)+log(y)⇒x=a , y=1+(b/a) ⇒  log(a+b)=log(a)+log(1+b/a)  ........  b=10^B  , a=10^A  , c=10^C   ((b/c))^(loga) =(10^(B−C) )^A =10^(BA−CA) ...  ⇒10^(BA−CA) ×10^(CB−AB) ×10^(AC−BC) =10^0 =1
$${log}\left({xy}\right)={log}\left({x}\right)+{log}\left({y}\right)\Rightarrow{x}={a}\:,\:{y}=\mathrm{1}+\frac{{b}}{{a}}\:\Rightarrow \\ $$$${log}\left({a}+{b}\right)={log}\left({a}\right)+{log}\left(\mathrm{1}+{b}/{a}\right) \\ $$$$…….. \\ $$$${b}=\mathrm{10}^{{B}} \:,\:{a}=\mathrm{10}^{{A}} \:,\:{c}=\mathrm{10}^{{C}} \\ $$$$\left(\frac{{b}}{{c}}\right)^{{loga}} =\left(\mathrm{10}^{{B}−{C}} \right)^{{A}} =\mathrm{10}^{{BA}−{CA}} … \\ $$$$\Rightarrow\mathrm{10}^{{BA}−{CA}} ×\mathrm{10}^{{CB}−{AB}} ×\mathrm{10}^{{AC}−{BC}} =\mathrm{10}^{\mathrm{0}} =\mathrm{1} \\ $$
Commented by mathlove last updated on 19/Apr/23
tnks
$${tnks} \\ $$

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