The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Exponents and logarithms (white questions)

After spending all day on this chapter, these are the questions I can't find the answers to after multiple rechecking.

I'd appreciate any help at all.

14553877560981456181527.jpg
(2D) (5) (a) (i)

*more to come*
1455387857972-1063140969.jpg525.7KB
1455387915300-81352283.jpg521.2KB
14553880969751497178341.jpg

(2E) (2) (e) (ii)

Need to express in terms of x, y and z

When
X = log a
Y = log b
Z = log c
1455388243613-1576788125.jpg

2E 3 e i

Need to find x
1455388348199-1493659826.jpg
2G 1 d i
1455388424494-759069640.jpg
The two I got wrong.
Reply 5
Avatar for TeeEm
TeeEm
19
what are white questions?
Original post by TeeEm
what are white questions?


The questions in my book are colour coded as white, yellow, green, blue and red :tongue:
Reply 7
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland

Spoiler



Why have you crossed out 2^32? It's correct.
Reply 8
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland
1455388243613-1576788125.jpg

2E 3 e i

Need to find x


You said log⁡3(expanded quadratic)=2\log_3 (\text{expanded quadratic}) = 2, so to get rid of the log, you need to do 323^2, not 232^3.
Reply 9
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland
1455388348199-1493659826.jpg
2G 1 d i


When you divide by 3, at the very end (3x = blah, to x = blah/3), you need to multiply the entire denominator by 3. That is:

log⁡37−3log⁡32−1×13=log⁡37−33(log⁡32−1)=log⁡37−33log⁡32−3\displaystyle \frac{\log_3 7 - 3}{\log_3 2 - 1} \times \frac{1}{3} = \frac{\log_3 7 - 3}{3(\log_3 2 - 1)} = \frac{\log_3 7 - 3}{3\log_3 2 - 3 }
Reply 10
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland
1455388424494-759069640.jpg
The two I got wrong.


You have 2xln⁡3=ln⁡5\displaystyle 2x \ln 3 = \ln 5, which I agree with, now divide both sides by ln⁡3\ln 3 to get:

2xln⁡3ln⁡3=ln⁡5ln⁡3  ⟺  2x=ln⁡5ln⁡3\displaystyle \frac{2x \ln 3}{\ln 3} = \frac{\ln 5}{\ln 3} \iff 2x = \frac{\ln 5}{\ln 3}

You seem to have done: 2xln⁡3=ln⁡5  ⟺  2x=ln⁡3ln⁡5\displaystyle 2x \ln 3 = \ln 5 \iff 2x = \frac{\ln 3}{\ln 5}, but that's not true/correct, can you see why?
Reply 11
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland
1455388424494-759069640.jpg
The two I got wrong.


Whilst we have the rule that a+bc=ab+ac\frac{a +b}{c} = \frac{a}{b} + \frac{a}{c}, it is not true that ab+c=ab+ac\frac{a}{b+c} = \frac{a}{b} + \frac{a}{c}! Test this out for yourself: 11+1≠11+11=2≠11+1=12\frac{1}{1+1} \neq \frac{1}{1} + \frac{1}{1} = 2 \neq \frac{1}{1+1} = \frac{1}{2}.

So whilst your ln⁡2ln⁡5−ln⁡2\displaystyle \frac{\ln 2}{\ln 5 - \ln 2} is correct, that's your final answer and you should leave it as such!
Reply 12
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland
The questions in my book are colour coded as white, yellow, green, blue and red :tongue:


Have I missed anything? :smile:
Thank you so much for your patience and help :woo:




Original post by IFoundWonderland
14553880969751497178341.jpg

(2E) (2) (e) (ii)

Need to express in terms of x, y and z

When
X = log a
Y = log b
Z = log c


Original post by Zacken
Have I missed anything? :smile:

Just this one :colondollar:

Original post by Zacken
Why have you crossed out 2^32? It's correct.

The book says it's wrong :dontknow:

Also, for this one, I corrected where you said, but still got the wrong answer :/ I got a negative but the answer wants a positive (I'm guessing 2 should be negative and 8 should be positive but idk how to get this)
1455395009531-1306395360.jpg
Reply 14
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland
Thank you so much for your patience and help :woo:


No problemo! Give me a sec, just wrapping some stuff up and I'll answer the rest of your questions in ~30 minutes! :smile:
Reply 15
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland

Also, for this one, I corrected where you said, but still got the wrong answer :/ I got a negative but the answer wants a positive (I'm guessing 2 should be negative and 8 should be positive but idk how to get this)
1455395009531-1306395360.jpg


You have x2−6x−16x^2 - 6x -16, but you've written: x2−2x+8x−16x^2 - 2x + 8x -16, note the two middle terms are −2x+8x=+6x-2x + 8x = +6x. you need it to be −6x-6x, that is, you need:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}x^2 + 2x - 8x - 16\end{equation*}



Pulling out −1-1 as a common factor in the last two terms:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}x^2 + 2x - (8x + 16) \end{equation*}



Pulling out 88 as a common factor from the brackets (note that you could have done the above step and this one simultaneously by pulling out −8-8)

Unparseable latex formula:

\displaystyle [br]\begin{equation*}x(x+2) - (8(x+2)) = x(x+2) - 8(x+2) = \cdots \end{equation*}

can you take it from here?
(edited 8 years ago)
Original post by Zacken
You have x2−6x−16x^2 - 6x -16, but you've written: x2−2x+8x−16x^2 - 2x + 8x -16, note the two middle terms are −2x+8x=+6x-2x + 8x = +6x. you need it to be −6x-6x, that is, you need:

x2+2x−8x−16=x2+2x−(8x+16)=x(x+2)−(8(x+2))=x(x+2)−8(x+2)=⋯\displaystyle x^2 + 2x - 8x - 16 = x^2 + 2x - (8x + 16) \\ = x(x+2) - (8(x+2)) = x(x+2) - 8(x+2) = \cdots can you take it from here?

Ahh yes, of course :facepalm:

Thank you so much for all your help - honestly don't have the words to express my gratitude.

You're an absolute superstar :h:
Reply 17
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland
Ahh yes, of course :facepalm:

Thank you so much for all your help - honestly don't have the words to express my gratitude.

You're an absolute superstar :h:


Edited my post to make it cleared! :smile:

Don't thank me just yet! I've got two more of your questions to reply to. :biggrin:

Protip: if you can't seem to factorise something correctly, it's worth just using the quadratic equation to find the roots in an exam or something.
Reply 18
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland


The book says it's wrong :dontknow:


You're correct, the book is wrong! :h:
Reply 19
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland

Just this one :colondollar:


Aaah, you're almost very nearly there! Note that you have:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\log 5 + y + \log 2 + 2z\end{equation*}



Shiftin things around:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\log 5 + \log 2 + y + 2z\end{equation*}



Using the addition -> multiplication rule thingy for logs.

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\log (5\times 2) + y + 2z = \log 10 + y + 2z\end{equation*}



Using the fact that log⁡10=log⁡1010=1\log 10 = \log_{10} 10 = 1 (in this case):

Unparseable latex formula:

\displaystyle [br]\begin{equation*}1 + y + 2z\end{equation*}



Note that we've used the rule log⁡ab+log⁡ac=log⁡abc\log_a b + \log_a c = \log_a bc with a=10a=10, b=5b=5 and c=2c=2.
(edited 8 years ago)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you