log6 4 + log6 9 (the 6's are the little numbers lower than the others.) how does this simplify and why?

Answer 1

2

when adding logs with similar base you multiply; 4*9 = 36
log (6) 36 = 2

Answer 2

well if you have two logs with the same bases (the little numbers as you like to put it) then the number that the base is to (im not sure of the actual term but your 4 and 9) are multiplied.

So the law goes like this logxy + logxz = logx(y+z)(where x is your base or your little number and y and z are two non-negative reals).

so the answer would be log6 (36) or log based 6 of 36

Answer 3

log (10) + log (1000) = log (10,000)
(1)+(3)=(4)

This is merely to show proof of the theorem above that log(a) + log (b) = log (ab). You can generalize this to include any base of the log.

now, why is it so? because the logarithmic scale is something of an inverse of the exponential scale. that is, it goes up slowly rather than very quickly. I wish I remembered more about logs to give you a better answer, but it's been a long time.

-edit-

good job Kevin! I knew taking antilogs was the answer, I was just braintired.

Answer 4

Suppose
log_b m + log_b n

Let x = log_b m
y = log_b n

thus,
b^x = m
b^y = n

Multiply the equations:
(b^x)(b^y) = mn

Add exponents
b^(x + y) = mn

Convert into log form
x + y = log_b mn

Substitute x and y
log_b m + log_b n = log_b mn

Your
log_6 4 + log_6 9
= log_6 (4)(9)
= log_6 36
= log_6 6²
= 2

^_^

Answer 5

log(6) 4+ log(6) 9= log(6) [489]=log(6) 36 = 2

Answer 6

log(6)4 + log(6)9 = log(6)(4 * 9) = log(6)36 = (log(36))/(log(6)) = 2

Answer 7

utilizing sources loga b= (log b)/ (log a) log3 4.log4 5.log5 6.log6 7.log7 8.log8 9 = (log4 .log5 .log6 .log7 .log8 .log9)/ (log3. log4 .log5 .log6 .log7 .log8 ) =log9/log3 =log3 9 =2