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Evaluate each expr… - QuestionCove
OpenStudy (anonymous):

Evaluate each expression. . .

6 years ago
OpenStudy (anonymous):

\[\log_{2}(1\32) \]

6 years ago
OpenStudy (anonymous):

the answer is -5

6 years ago
OpenStudy (angela210793):

\[\log_{2} 2^{-5}=-5\log_{2} 2=-5*1=-5\]

6 years ago
OpenStudy (angela210793):

cause 1/32=32^-1=2^-5

6 years ago
OpenStudy (anonymous):

okaay, you lost me with the -5 * 1? How did you get that?

6 years ago
OpenStudy (angela210793):

\[\log_{a} b ^{c}=c*\log_{a}b \]

6 years ago
OpenStudy (angela210793):

the exponent goes in front of the log and multiplies it

6 years ago
OpenStudy (anonymous):

I understood that part but how does \[\log_{2}2 \] = 1?

6 years ago
OpenStudy (angela210793):

and \[\log_{a}a=1 so \log_{2} 2\]

6 years ago
OpenStudy (angela210793):

yea

6 years ago
OpenStudy (anonymous):

I stilll don't understand. :( What is \[1solog_{2}2?? \]

6 years ago
OpenStudy (angela210793):

\[\log_{a} a=1\]

6 years ago
OpenStudy (anonymous):

Oh, always?

6 years ago
OpenStudy (angela210793):

yea...\[\log_{58} 58=1\]

6 years ago
OpenStudy (anonymous):

and what if it is \[\log_{a}b? \]

6 years ago
OpenStudy (angela210793):

then \[\log_{a} b=c\]

6 years ago
OpenStudy (angela210793):

so c^a=b

6 years ago
OpenStudy (angela210793):

ex: \[\log_{2} 32=5 \]

6 years ago
OpenStudy (angela210793):

Because \[2^{5}=32\]

6 years ago
OpenStudy (anonymous):

but isn't it c^a=b? your example looks like a^c=b

6 years ago
OpenStudy (angela210793):

sorry it is a^c..sorry..typing mistake :(:(

6 years ago
OpenStudy (anonymous):

its okay no worriess:p Okay I get it now, i think. Haha. Thank youu!

6 years ago
OpenStudy (angela210793):

I really hope u do :):) and ur welcome :)

6 years ago
OpenStudy (anonymous):

I'm sorry to bother you again but I figured you're good with logs:p Solve each equation: \[\log_{7}9+\log_{7}-2x=\log_{7}19 \] The answer is: \[\left\{ (-19\18) \right\}\]

6 years ago
OpenStudy (angela210793):

It's ok i like them :)

6 years ago
OpenStudy (anonymous):

haha, i knew it! :p

6 years ago
OpenStudy (angela210793):

log_7 of -2x??????

6 years ago
OpenStudy (anonymous):

yeep.

6 years ago
OpenStudy (anonymous):

okaay. that's weirdd.

6 years ago
OpenStudy (angela210793):

w8w8w8w8..let me see again :)

6 years ago
OpenStudy (angela210793):

u want to find X right?

6 years ago
OpenStudy (angela210793):

there's a rule that says : \[\log_{a} b+\log_{a}c=\log_{a} b*c\]

6 years ago
OpenStudy (angela210793):

so \[\log_{7}9+\log_{7} -2x=\log_{7} 19\]

6 years ago
OpenStudy (anonymous):

I think? I'm so confused. . .

6 years ago
OpenStudy (angela210793):

\[=\log_{7} [9*(-2x)]=\log_{7} 19\]

6 years ago
OpenStudy (angela210793):

\[\log_{7}(-18x)=\log_{7} \]

6 years ago
OpenStudy (angela210793):

since the bases ar the same we can equal -18x=19 therefore x=-19/18

6 years ago
OpenStudy (anonymous):

ohh...... you're amazing you know that?! hahaha

6 years ago
OpenStudy (anonymous):

you've saved my retricetwice. Congrats(y)

6 years ago
OpenStudy (angela210793):

Hahaha loool :P:P no I'm not...:P

6 years ago
OpenStudy (anonymous):

yess you aree. thank you sooo muchh(Y)

6 years ago
OpenStudy (angela210793):

Uw :) (ur not cheating in a test r u?)

6 years ago
OpenStudy (anonymous):

No!!! Haha, I have finals-.-

6 years ago
OpenStudy (angela210793):

oh...good to hear tht...hmm..let me see if i can give u a link abt log

6 years ago
OpenStudy (anonymous):

what if it ends up being \[2x ^{2}-6=44?\]

6 years ago
OpenStudy (anonymous):

Thanks!

6 years ago
OpenStudy (angela210793):

http://en.wikipedia.org/wiki/Logarithm

6 years ago
OpenStudy (angela210793):

2x^2=44+6 2X^2=50 x^2=50/2 x^2=25 x=+-5

6 years ago
OpenStudy (anonymous):

ahh. okaay, correction, 3 timesss(y)

6 years ago
OpenStudy (anonymous):

And thanks for the website:p

6 years ago
OpenStudy (angela210793):

ur welcome :) Sometimes it is easier to learn smth from internet than books :P:P

6 years ago
OpenStudy (anonymous):

Yeah, that's truee.But what about natural logs? I have one here that says \[\ln 3+\ln (x-8)=3\] and i was able to get all the way up to \[\ln 3x-24=3\] but i dont know what to do from there.

6 years ago
OpenStudy (anonymous):

huh?

6 years ago
OpenStudy (angela210793):

yea..remember when i told u tht log_a,b^c=c*log_a,b?

6 years ago
OpenStudy (anonymous):

yesss.

6 years ago
OpenStudy (angela210793):

so 3 is an exponent ...and ln n is=1 so we can say that 3=3*ln n

6 years ago
OpenStudy (anonymous):

wait but how did you get 3ln n from 3?

6 years ago
OpenStudy (anonymous):

the answer is \[\left\{ (e ^{3}+24)/3 \right\}\]

6 years ago
OpenStudy (angela210793):

i explained that above..... u know that 3=3*1 right...so 1=ln e in this case we must have log with same base (*e in this case) so we can equal 3x-28

6 years ago
OpenStudy (anonymous):

Okay. so ln=e?

6 years ago
OpenStudy (angela210793):

hmnmhmhm..can i write it again plsss???????????????? I've done some very stuuuuuuuuupid mistakes

6 years ago
OpenStudy (anonymous):

haha, its okay, go ahead. break it down please. I'm kinda stupid on the weekends:p

6 years ago
OpenStudy (angela210793):

let me delete all those and I'll explain better

6 years ago
OpenStudy (angela210793):

ln(3x-24)=3 ln(3x-24)=3*ln e ln(3x-24)=lne^3 so base is the same=e and we can equal 3x-24=e^3 we add 24 from both parts 3x=e^3+24 then we divide with 3 x=(e^3+24)/3 the same answer as u have :) got it? PS: I'm so sorry that i messed u up....i've been studying math all day so..my brain is a bit ''out of order''

6 years ago
OpenStudy (anonymous):

Lol, okay I think I get it now, I'm gonna try to do it by myself and see if i get it. And don't worrry about, I'm out of order too:p

6 years ago
OpenStudy (angela210793):

K ^_^ :) LOL :D

6 years ago
OpenStudy (anonymous):

I get it! YAY! :D

6 years ago
OpenStudy (anonymous):

Okay, new problem for something completely different or am I boring you and you wanna go sleep?

6 years ago
OpenStudy (angela210793):

Happy for you :)(it's 4:13 pm here...no sleep at this time...) Ur not boring me at all...is just that idk if i'll know how to answer...u ask it anyway

6 years ago
OpenStudy (anonymous):

Okay, rofl, its 10 in the morning here, I'm super tired and I have tutoring in an hour-.- okay, next problem. Solve each equation. Round your answers to the nearest ten-thousandth. \[3\times19^{1.8x +7}=65\]

6 years ago
OpenStudy (angela210793):

10 am and ur tired?????????? oO is tht x multiplying or as a variable?

6 years ago
OpenStudy (anonymous):

yes, 10 am and im exhausted. I normally don't wake up until 12:p and i really don't know:p

6 years ago
OpenStudy (angela210793):

Hahah lool...I wake up at 7am :P Come on...how u don't know?? Oo

6 years ago
OpenStudy (anonymous):

Oh wow. Early riserr. And I don't know the difference. But if I had to guess I would say variable. The answer is -3.3086

6 years ago
OpenStudy (angela210793):

:P:P:P hmmmmmm...let me work on a paper 1st and see if i can solve it ...

6 years ago
OpenStudy (anonymous):

Okaay, you do that and I'll be right back:p

6 years ago
OpenStudy (angela210793):

[19^(1.8x+7)]*x=65/3 \[\log_{19} 19^{1.8x+7} * x=\log_{19}65/3 \]

6 years ago
OpenStudy (angela210793):

\[(1.8x+7)=\log_{19}65-\log_{19}3-\log_{19}x \]

6 years ago
OpenStudy (angela210793):

:(:(:( I'm sorry....Idk how to transform it......:(:(:( I'm really sorry :(:(

6 years ago
OpenStudy (angela210793):

It'd b easier if X was ''multiply''

6 years ago
OpenStudy (angela210793):

Can u please think if it really is a variable?

6 years ago
OpenStudy (anonymous):

its okay dont worry about itt:p

6 years ago
OpenStudy (angela210793):

U know...i was thinking u might have gone to bed :P:P Sorry and thnx :)

6 years ago
OpenStudy (anonymous):

Noo, hahaha! But I am going to tutoring now:p Thanks for the helpp! TTYL

6 years ago
OpenStudy (angela210793):

Ur welcome... Have a gr8 time,see ya :)

6 years ago
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