'animechic' i dont think you can just divide by log 2. you have to know what opperations there are and then apply the log rules for them, this is what i mean:
well 'Goldfinger' was right to say that the first thing to do is to remove cofficients:
log[2]16 + log[2]4^3 - log[2] 8
now from BODMAS we know that in this case we should add 1st, and we know from our log rules that when adding common logs we multiply:
log[2](16*64) - log[2] 8
we also know from our log rules that when subtracting common logs we divide:
no 4 for this problem the same principles apply as in no 2. these are 1)remove cofficients 2)when adding common logs, multiply 3)when subtracting common logs, divide.
i think the equation is equal to 3.4, it looks like that so im using that, correct me if its something else:
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problem 1:
ReplyDeletelog [3] 7 + log [3] 5
first you check to see if the logs are of the same base. if they are not of the same base, you make them both base 10.
always remember that log[2] 8 = 3 is the same as 2^3 = 8,
log a + log b = log a*b,
log a - log b = log a/b.
with those in mind you can appraoch to make the log question into one expression.
log[3] 7*5
can #2 be worked out by dividing the equation by log 2
ReplyDeleteto answer your question "animechic"
ReplyDeleteyes u can
but first you have to get rid off the coefficient
making the equation
log[2]16 + log[2]4^3 - log[2]8
hope this help
problem 3
ReplyDeletefirst there are no coefficient
we notice the bases are the same
so
using the log rule
log[a]b + log[a]c = log[a]bc
the equation become
log[5](9)(24)
then using
rule log[a]b = logb/loga
we get log(9*24)/log5
we solve for log
then divide
to get our answer
no 2
ReplyDelete'animechic' i dont think you can just divide by log 2. you have to know what opperations there are and then apply the log rules for them, this is what i mean:
well 'Goldfinger' was right to say that the first thing to do is to remove cofficients:
log[2]16 + log[2]4^3 - log[2] 8
now from BODMAS we know that in this case we should add 1st, and we know from our log rules that when adding common logs we multiply:
log[2](16*64) - log[2] 8
we also know from our log rules that when subtracting common logs we divide:
log[2](1024/8)
log[2] 128
no 4
ReplyDeletefor this problem the same principles apply as in no 2. these are 1)remove cofficients 2)when adding common logs, multiply 3)when subtracting common logs, divide.
i think the equation is equal to 3.4, it looks like that so im using that, correct me if its something else:
log[5] (5*10) - log[5] 3 = 3.4
log[5] (50/3) = 3.4
log[5] 16.6 = 3.4
log [3] 7 + log [3] 5
ReplyDeletelog [3] (7*5)
log [3] 35
Darky don't you have to solve out the question?
ReplyDeleteOr sorry i now saw the question asked to express not solve.
ReplyDeleteQuestion : Let logb(2) = 0.3869, logb(3) = 0.6131, and logb(5) = 0.8982. Using these values, evaluate logb(10).
ReplyDeleteSince 10 = 2 × 5, then:
logb(10) = logb(2 × 5) = logb(2) + logb(5)
Since I have the values for logb(2) and logb(5), I can evaluate:
logb(2) + logb(5) = 0.3869 + 0.8982 = 1.2851
Then logb(10) = 1.2851.
Question : Let logb(2) = 0.3869, logb(3) = 0.6131, and logb(5) = 0.8982. Using these values, evaluate logb(9).
ReplyDeleteSince 9 = 32, then:
logb(9) = logb(32) = 2logb(3)
Since I have the value for logb(3), then I can evaluate:
2logb(3) = 2(0.6131) = 1.2262
Then logb(9) = 1.2262.
QUESTION TO DO:
ReplyDeleteLet logb(2) = 0.3869, logb(3) = 0.6131, and logb(5) = 0.8982. Using these values, evaluate logb(7.5).
1.1244
Deletequestion 1
ReplyDeletelog[3] 7 + log[3] 5
we realize they have sme base so therefore the answer will be:
log[3]7*5
in no. 2 we have a 3 infront of the log so that indicates that we have to remove the 3 and put it as a power
ReplyDeleteso the quetion will now read:
log[2]16 + log[2]4*3 - log[2]8
then we get 4 + 6 - 3 = 7
again in question 3, we have same bases so
ReplyDeletelog[5]9 + log[5]24
and the opperation in addition so we get
log[5]9*24
Question 1
ReplyDeletelog [3] 7 + log [3] 5
they are the same bases so we multiply therefore we get
log [3] (7*5)
log [3] 35
log 35/ log 3 = 1.46
Question 2
ReplyDeletelog [2] 16 + 3log [2] 4 - log [2] 8
firstly we work out the addition part of the question and sine the have the same base we multiply and we get
log [2] 16 + log [2] 4^3 - log [2] 8
log [2] 16 + log [2] 64 - log [2] 8
log [2] (16*64) - log [2] 8
log [2] 1024 - log [2] 8
Now since it is subtraction and has the same base we divide and we get
log [2] 1024 / 8
log [2] 128
log 128 /log 8 = 7
Question 3
ReplyDeletelog [5] 9 + log [5] 24.
The base are the same so we multiply because its addition
log [5] (9 * 24)
log [5] 216
log 216 / log 5 = 3.34
For the log questions above always remember that when you are add two logs on with the same base, you multiply and when substracting, you devide...
ReplyDelete1)log [3] 7 + log [3] 5
ReplyDeleteapply the rules of logs
log a + log b = log a*b
log 3 35
3)log [5] 9 + log [5] 24
ReplyDeleteapply the rules of logs
log a + log b = log a*b
log 5 24*9
log [2] 16 + 3log [2] 4 - log [2] 8
ReplyDeletein this question we are applying the rules of logs
when adding and subtracting
first thing remove all coefficients
log [2] 16 + log [2] 64 - log [2] 8
log [2] (16*64) - log [2] 8
log [2] 1024 - log [2] 8
now we apply the subtracting rule so we divide.
log [2] 1024 / 8
log [2] 128
calculator dsent have base 2 so we use base convert to base 10
log 128/ log 2 =