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Question-208384

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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) (a)+log_(abc) (b)=2+3=5⇒log_(abc) (ab)=5 log_(abc) (abc)=1=log_(abc) (ab)+log_(abc) (c)=5+? ⇒?=1−5=−4
logabc(a)+logabc(b)=2+3=5⇒logabc(ab)=5logabc(abc)=1=logabc(ab)+logabc(c)=5+?⇒?=1−5=−4
Answered by Rasheed.Sindhi last updated on 14/Jun/24
{ ((log_(abc) (a)=2 )),((log_(abc) (b)=3 )),((log_(abc) (c)=? )) :}⇒ { (((abc)^2 =a)),(((abc)^3 =b)),(((abc)^x =c)) :} (abc)^(2+3+x) =abc 5+x=1⇒x=−4
{logabc(a)=2logabc(b)=3logabc(c)=?⇒{(abc)2=a(abc)3=b(abc)x=c(abc)2+3+x=abc5+x=1⇒x=−4

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