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Question Number 145406 by mathdanisur last updated on 04/Jul/21
if log_a c+log_c b=2 ; log_b c+log_a c=0 find (1/(log_a b)) + (1/(log_b c)) + (1/(log_c a)) = ?
iflogac+logcb=2;logbc+logac=0find1logab+1logbc+1logca=?
Answered by som(math1967) last updated on 04/Jul/21
log_b c+log_a c=0 log_b c×log_c b=−log_a c×log_c b 1=−log_a b ∴log_b a=−1 (1/(log_a b))+(1/(log_b c))+(1/(log_c a)) log_b a+log_c b+log_a c −1+2=1 [∵log_a c+log_c b=2]
logbc+logac=0logbc×logcb=−logac×logcb1=−logab∴logba=−11logab+1logbc+1logcalogba+logcb+logac−1+2=1[∵logac+logcb=2]
Commented by mathdanisur last updated on 04/Jul/21
alot cool thankyou Ser
alotcoolthankyouSer
Answered by liberty last updated on 05/Jul/21
{ ((log _a c+log _c b = 2 ...(1))),((log _b c + log _a c = 0 ...(2))) :} we know that log _a c = (1/(log _c a)) so log _b c + log _a c = 0 ⇒(1/(log _c b)) = −(1/(log _c a)) or log _c a =−log _c b then a = (1/b) ...(3) substract eq(1) &(2) give ⇒ log _c b−log _b c = 2 now consider that (1/(log _a b)) + (1/(log _b c)) + (1/(log _c a)) = log _b a + log _c b + log _a c = −1+ log _c b−log _b c = −1+2 = 1.
{logac+logcb=2…(1)logbc+logac=0…(2)weknowthatlogac=1logcasologbc+logac=0⇒1logcb=−1logcaorlogca=−logcbthena=1b…(3)substracteq(1)&(2)give⇒logcb−logbc=2nowconsiderthat1logab+1logbc+1logca=logba+logcb+logac=−1+logcb−logbc=−1+2=1.
Commented by mathdanisur last updated on 05/Jul/21
alot cool thanks Ser
alotcoolthanksSer

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