Evaluate: log(5^3) (17^6)
a) 2 log(12^5) (17)
b) 0.5 log17
c) log5 (17^18)
d) log5 (17)
e) 2 log5 (17)

• here first of all, (5^3) is a base for log (17^6) please dont think as log(a)(b) where a= 5^3 and b=17^6.
  Next thing is to make equal base we have a formula as logb (a)= logc (a)/ logc (b).
  now after evaluating it will be 6log(5^3) 17, then applying formula as 6 log5 (17)/ log5 (5^3) where a=17, b=(5^3) and c =5.
  then 6 log5 (17)/ 3log5 (5)= 2 log5 (17)/ log5 (5)= 2log5 (17). OPTION (e).
  THAAAAAAANKKSSSSSS
• 5 years agoHelpfull: Yes(9) No(5)
• logab=logb/loga;
  =6log17/3log5
  =2log(5)(17) (e)
• 3 years agoHelpfull: Yes(1) No(0)