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AR15.COM
4/27/2011 1:35:54 PM EDT


Thanks AllenNH

4/27/2011 1:39:03 PM EDT
[#1]
Try wiki
_______________________________
("Captain's log, stardate.....,"––Kirk, (w,stte), "ST:TOS")
4/27/2011 1:39:10 PM EDT
[#2]

In before 288.


4/27/2011 1:40:03 PM EDT
[#3]




Quoted:

Try wiki

_______________________________

("Captain's log, stardate.....,"––Kirk, (w,stte), "ST:TOS")




I understand how to work these problems, just not this one.



It has me stumped
4/27/2011 1:40:36 PM EDT
[#4]



Quoted:




In before 288.











2 damn it



 
4/27/2011 1:41:36 PM EDT
[#5]
Quoted:

Quoted:

In before 288.





2 damn it
 


+1+1 = 2  
4/27/2011 1:42:37 PM EDT
[#6]
x = 87
4/27/2011 1:42:50 PM EDT
[#7]
thought thread said Meth experts. I am disappoint.
4/27/2011 1:51:04 PM EDT
[#8]
Quoted:
I know this is just College algebra, but for the life of me I cant figure this question out. I have a online quiz and this guestion came up
Solve:
https://assessment.casa.uh.edu/Assessment/Images//1225837.gif

I got x=11, which is wrong. I cannot figure out what I am doing wrong.

Any help would be much appreciated, thanks



For serious?

That simplifies to log11(x) = 10.

So, what power must 11 be raised to, to = 10?

(p.s.  288)

ETA: Oh since you're 'stuck', I'll show my work.

original:
121 = 10Log11(11x) + 11

121 - 11 = 10Log11(11x)
(121 - 11)/10 = Log11(11x)
110/10 = Log11(11x)
11 = Log11(11x)
11 = Log11(x) + log11(11)
11 - log11(11) = log11(x)
11 - 1 = log11(x)
10 = log11(x)
4/27/2011 1:59:14 PM EDT
[#9]




Quoted:



Quoted:

I know this is just College algebra, but for the life of me I cant figure this question out. I have a online quiz and this guestion came up

Solve:

https://assessment.casa.uh.edu/Assessment/Images//1225837.gif



I got x=11, which is wrong. I cannot figure out what I am doing wrong.



Any help would be much appreciated, thanks







For serious?



That simplifies to log11(x) = 10.



So, what power must 11 be raised to, to = 10?



(p.s. 288)



ETA: Oh since you're 'stuck', I'll show my work.



original:

121 = 10Log11(11x) + 11



121 - 11 = 10Log11(11x)

(121 - 11)/10 = Log11(11x)

110/10 = Log11(11x)

11 = Log11(11x)

11 = Log11(x) + log11(11)

11 - log11(11) = log11(x)

11 - 1 = log11(x)

10 = log11(x)



Thank you. Im sorry if Im retarded.



makes sense now that i see it and know how to do that for the future.



Im not sure what I was thinking, I already worked multiple problems out, then derped all over this one.



I was trying to use the inverse operation on the log11(11x) to get just x......
4/27/2011 1:59:27 PM EDT
[#10]
maybe put in these terms you'll understand:

11 + 10v11(11x) = 121
4/27/2011 2:00:21 PM EDT
[#11]








cool



No need for the insult, but thank you for trying to help



I had a derp moment and now understand how I herped.
4/27/2011 2:29:22 PM EDT
[#12]
How many "11's" (the base) have to be multiplied together to get the answer after rearranging the equation?

log(base 11) (11X) = 11

tilt
4/27/2011 2:40:14 PM EDT
[#13]
Quoted:



cool

No need for the insult, but thank you for trying to help

I had a derp moment and now understand how I herped.


No insult intended, sorry you took it that way.

ETA: Shake your mangina, its the only way to get ALL of the sand out!
4/27/2011 2:41:26 PM EDT
[#14]
Per OP