Question

evaluate log 6 + 2log5+log 4 -log 3 -log 2

Answer 1

hello friends......

solution:-
using formulas
log m×n= log m + log n

=>log 6 + 2log5+log 4 -log 3 -log 2 =
log 2×3 + 2log 5 + log 2×2 - log 3 -log 2
=> log 2 + log 3 +2 log 5 +log 2+ log 2 - log 3 - log 2
=> 2log 2 +2 log 5 = 2( log 2 + log 5)
=> 2log 2×5 = 2log 10 = 2 answer

♦♦ hope it helps ♦♦

Answer 2

HELLO MATE ! the solution for your question is evaluated by the formula

log m*n=log m + log n

Step-by-step explanation:

log6 +2log5+log4-log3-log2

⇒log2+log3 +2log5+log2+log2-log3 -log2 (simplifying into small parts)    

⇒ 2log2 + 2log5                                                                            

⇒2(log2+log5)                                                     (taking common)

⇒2(log2×5)

⇒2log10

⇒2×1                                                                     (∵  log10=1)      

⇒2

     hope this answer helps you!!