how do you write this logarithmic expression as a single logarithm : 5 log 3 + log 4
log 3^5 + log 4 = log (243*4)
does t make sense? log (a * b) = log a + log b and n log x= log x^n
yea it does make sense , thank you
:)
bt are u sure it is right?
because isnt it supossed to be like similar to this problem : log 7 + log 2 which equals log 3.5 because u have to divide 7/ 2 .... right ?
@akash123
log (a * b) = log a + log b so log 7 + log 2 = log (7 * 2)= log 14
log 7 - log 2 = log(7/2) = log 3.5 since log a - log b = log(a/b)
wait i dont get it ... why did u just change ur answer from multiplying to dividing the problem ?
log ( a *b)= log a+ log b log(a/b) = log a - logb
in the beginning, it was a typo but i changed it at that time only
sorry...now it's fine?
5 log 3 + log 4= log 3^5 + log 4 = log (243*4)
wait i still dont get it so u multiply that problem instead of dividing ?
arent u supposed to divide?
@akash123
no...log a + log b = log (a*b)
oh okay thank you , bt why did u put 3^5 instead of putting 5^3
?
since n log x= log (x^n) so 5 log 3= log (3^5) = log 243
oh okay thank you
so how do u solve log 8 - 2 log 6 + log 3
?
log 8 + log 3 - 2 log 6 = log( 8*3) - log (6^2) = log 24 - log 36 = log(24/36)
okay i have a couple more questions : how would you write these logarithmic expressions as a single logarithm : 1. 4 log m - log n 2. log 5 - k log 2 3.) log [ base of 6] 5 + log [ base of 6 ] x 4. log { base of 7 ] x + log [base of 7] y - log [base of 7] z
try 1st Q by ur own ...i'll help u
okay would the answer for the first one be .... log 4^m + log n = log 4/mn
4 log m = log m^4...right?
ye i think so
4 log m - log n = log m^4/n
oh okay i get it
so what is the answer for the ne
next one *
?
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