Chocolates Questions

[Chocolate]

Looks like I'll follow the trend and make my own thread 

There are a couple of log questions i need help with..albeit them being *easy* ones.

1) Simplify Log3 6 + log3 x^1/2 = 1

I did..
Log3 6 + log3 X^1/2 = Log3 3
log3 X^1/2= Log3 3 - Log3 6
log3 X^1/2= log3 3/6
1/2log3 X= log3 3/6
log3 X = 2log3 3/6
log3 X = log3 (3/6)^2
log 3 X = log3 1/4                     [is cancelling the log3 out possible?]
X = 1/4

and the answers have 1/4 as the answer...how did they rid the equation of the x? (noting that, if correct, it should say x=1/4 as the answer because they are simplifying NOT solving..) Or did they just leave it out..?

ok and next one

2)
(2^x) . (3^x+1) = 10

I wont type it all, I'll just explain what I did.
Brought the x to the front and made xlog10 2
Split the 3^x+1 into (3^x . 3)
Logged both of these
Multiplied these with xlog10 2
Logged 10
Divided then squared..
aaand
got the wrong answer

Help appreciated.

[Mao]

hehe~ welcome

1)
yes, you can cancel logs on both side.

and the answers have 1/4 as the answer...how did they rid the equation of the x? (noting that, if correct, it should say x=1/4 as the answer because they are simplifying NOT solving..) Or did they just leave it out..?

and there's no need to worry about this kind of tiny detail
if we are to be pedantic:
expressions (without equal/inequal signs) are simplified, equations are solved in respect to a variable. so equations cannot really be simplified, but expressions within the equation can be.

2)
the way you wrote the question was a little ambiguous, but after reading your explanations this is what I think you meant:



which is:










it will be a great help if you could use the math typeset.
to use LaTeX:

Code: [Select]

[tex]....expression....[/tex]
a short tutorial on different mathematical expressions:
http://vcenotes.com/forum/index.php/topic,864.0.html

[Chocolate]

*hits head* How could I not realise that I had to group the ^x's...Silly silly

Thankyou very much Mao 

[Chocolate]

I'm back with a few easy baby questions..that I can't do

[I'd use latex but I'm feeling lazy..sorry, the expressions look terrible :angel:]


Simplify:

5^(3n) x 16^(n) / 50^(2n) x 4^(-n)   x  (16/125)^n 

I keep getting..

5^(n) x 16.(4)^(3n) / 50.(5)^(n+1)

I sub back n=1 to see if they give the same answer, and my simplified expression is a decimal point in difference..ie 0.4096 as opposed to my 4.096.. lol

and also
given that a= b^(c)

find b in terms of c and a
find c in terms of a and b
OK for the first one i tried log(e)'ing both sides
So  I went Log(e)a = Log(e)b^c
Log(e)a= c x Log(e)b
divide by log e
b= a/c
So very wrong.

and then for the second one, I just did
C= log(b)a
which I'm sure is alright.

[Mao]

grouping everything together"







(grrr that was so hard to read in terms of LaTeX code, haha)




1)




from your method, we could have done:







etc ~
and you are right for the second one =)

[Chocolate]

grouping everything together"







(grrr that was so hard to read in terms of LaTeX code, haha)




1)




from your method, we could have done:







etc ~
and you are right for the second one =)

Thanks Mao, very much appreciated, expect you made a little mistake on question 2..(I dont blame you, all that latex  :uglystupid2:)

you said


when the 2^(-2n) on the right should be 2^(2n) because you brought the negative up in order to multiply, so it becomes positive.

Then you get.. (256/625)n and then when you sub in 1 you get 0.4096 as you should. 

[Mao]

ooh, is that what the question asked?




sorry

[Chocolate]

No, no you're perfectly ok, I just didnt define the question properly. Look at it, lol

p.s. I'm taking you into my methods and spesh exams