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Question Number 84000 by byaw last updated on 08/Mar/20

Commented by Kunal12588 last updated on 08/Mar/20

log ((0.0005957))^(1/3) should be −ve ?

log0.00059573shouldbe−ve?

Answered by MJS last updated on 08/Mar/20

log_b 5.957 =.7750 ⇔ 5.957=b^(.7750) (((5.957)/(10000)))^(1/3) =b^x x=((ln (((5.957)/(10000)))^(1/3) )/(ln b))=(((1/3)(ln 5.957 −4ln 10))/(ln b))= =(1/3)(((ln 5.957)/(ln b))−((4ln 10)/(ln b)))=(1/3)(log_b 5.957 −4log_b 10)= =(1/3)(.7750−4log_b 10)= [b^(.7750) =5.957 ⇒ b≈10] =−1.075

logb5.957=.7750⇔5.957=b.77505.957100003=bxx=ln5.957100003lnb=13(ln5.957−4ln10)lnb==13(ln5.957lnb−4ln10lnb)=13(logb5.957−4logb10)==13(.7750−4logb10)=[b.7750=5.957⇒b≈10]=−1.075

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