Assignment Logarithms Question

[Dreyzo]

Hi guys, I’m just horribly stuck on this question, I’ve attempted 4a), but I’m not too sure if it is correct.
I’m very stuck on 4b), I don’t want to be fed the answer, but maybe a hint of what concept, or logarithm law I should be using.
Thanks in advance !!

 

[Deleted member 4993]

Hi guys, I’m just horribly stuck on this question, I’ve attempted 4a), but I’m not too sure if it is correct.
I’m very stuck on 4b), I don’t want to be fed the answer, but maybe a hint of what concept, or logarithm law I should be using.
Thanks in advance !!

It is admirable that you don't want to be fed the answer. But:

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[Dr.Peterson]

Hi guys, I’m just horribly stuck on this question, I’ve attempted 4a), but I’m not too sure if it is correct.
I’m very stuck on 4b), I don’t want to be fed the answer, but maybe a hint of what concept, or logarithm law I should be using.
Thanks in advance !!

Part (a) uses one of the standard graph transformations you will have learned previously.

Part (b) provides two pairs, (10,0) and (1,-2), that satisfy the equation. Plug those into the equation and solve for k and c.

Is that enough to get started? Show your work if you need more hints or corrections.

 

[Dreyzo]

Part (a) uses one of the standard graph transformations you will have learned previously.

Part (b) provides two pairs, (10,0) and (1,-2), that satisfy the equation. Plug those into the equation and solve for k and c.

Is that enough to get started? Show your work if you need more hints or corrections.

Hi, thanks for replying, for 4a), I assume it is a dilation of factor 1/2 in the y axis? I’ve drawn what I think the graph would look like in the photo.
For 4b) I’ve started to write a simultaneous equation, but I’m not too sure how to proceed. I’ve gone subtracting them from each other, so they end up in a fraction, but it doesn’t look too simplified to me. I’ll send the photo ASAP of my working—currently away from my desk.

 

[Dr.Peterson]

Hi, thanks for replying, for 4a), I assume it is a dilation of factor 1/2 in the y axis? I’ve drawn what I think the graph would look like in the photo.

The dilation is in the x (horizonal) direction, not "in the y axis"; maybe you meant "perpendicular to the y axis". And I didn't take your graph to be your answer because the factor there is not 1/2. The point that I now see you labeled as 5 is very inaccurate!

For 4b) I’ve started to write a simultaneous equation, but I’m not too sure how to proceed. I’ve gone subtracting them from each other, so they end up in a fraction, but it doesn’t look too simplified to me.

One equation should be very simple, when you evaluate the log, so you shouldn't need to subtract equations.

 

[Steven G]

k*log₂(2x) = k(log₂2 + log₂x) = k + k(log₂x)

 

[Dreyzo]

The dilation is in the x (horizonal) direction, not "in the y axis"; maybe you meant "perpendicular to the y axis". And I didn't take your graph to be your answer because the factor there is not 1/2. The point that I now see you labeled as 5 is very inaccurate!


One equation should be very simple, when you evaluate the log, so you shouldn't need to subtract equations.

Thanks for replying, would the dilation make the co-ordinate become (20,0) then?... I'm very confused, I need to familiarise myself with transformations in log graphs, it is currently school holidays and all information has left my brain.

In terms of question 4b), would you do the question like this?

- Substitute the co-rresponding k & c terms into the equation
- Re arrange to have one of the terms by itself
- and then re - subtitute to find the other term??

I'm so sorry if this is hopelessly nowhere near what I'm supposed to do. I thank you for your time.

 

[lex]

@Dreyzo
Your graph is correct, just the 5 is too close to the 10.
For part (b) using [MATH] \,\, y=k \log_2 x +c\hspace2ex[/MATH] (*)
(i) substitute the point (10,0), i.e. x=10, y=0 into the equation above (*)
(ii) substitute the point (1,-2), i.e. x=1, y=-2 into the equation above (*)
The second one gives you c directly, and the first will then allow you to find k.

 

[Dreyzo]

@Dreyzo
Your graph is correct, just the 5 is too close to the 10.
For part (b) using [MATH] \,\, y=k \log_2 x +c\hspace2ex[/MATH] (*)
(i) substitute the point (10,0), i.e. x=10, y=0 into the equation above (*)
(ii) substitute the point (1,-2), i.e. x=1, y=-2 into the equation above (*)
The second one gives you c directly, and the first will then allow you to find k.

Ah! Thank you so much, I realise now what to do, thank you. Is this correct? I have attached a photo of my working. Thanks for your time.

 

[Dr.Peterson]

Ah! Thank you so much, I realise now what to do, thank you. Is this correct? I have attached a photo of my working. Thanks for your time.

Almost orrect, but you should have much earlier replaced \(\log_2(1)\) with its value, 0! Then, as you were told twice, the second equation would immediately give the value of c. The work is a lot easier that way, and you would have avoided a sign error and an error in handling exponents. (Do you really think that \(2^{-2}=2^{1/2}\)?)

 

[lex]

As Dr.Peterson has pointed out:
You missed the fact that [MATH]c=-2[/MATH] (directly from the second equation, since [MATH] \log_2 1=0[/MATH]).
Incidentally you didn't answer the question asked, because you didn't find c.

There were two errors in the working:
you thought that [MATH]2^{-2}[/MATH] is [MATH]2^{1/2}[/MATH]and you got a sign wrong; there should be a line:
[MATH]0=k\log_2 10 - 2\hspace2ex[/MATH] (not +2) (and remembering that [MATH] \log_2 1=0[/MATH]).
so [MATH]k=\dfrac{2}{\log_2 10}[/MATH]
Answer: [MATH]c=-2, k=\dfrac{2}{\log_2 10}[/MATH]
(Incidentally, the original expression can simplify to [MATH]y=2(\log_{10} x - 1)[/MATH]).

 

[Dreyzo]

Almost orrect, but you should have much earlier replaced \(\log_2(1)\) with its value, 0! Then, as you were told twice, the second equation would immediately give the value of c. The work is a lot easier that way, and you would have avoided a sign error and an error in handling exponents. (Do you really think that \(2^{-2}=2^{1/2}\)?)

Thank you !!! I realise my careless mistakes. Thank you again for your time.

 

[Dreyzo]

As Dr.Peterson has pointed out:
You missed the fact that [MATH]c=-2[/MATH] (directly from the second equation, since [MATH] \log_2 1=0[/MATH]).
Incidentally you didn't answer the question asked, because you didn't find c.

There were two errors in the working:
you thought that [MATH]2^{-2}[/MATH] is [MATH]2^{1/2}[/MATH]and you got a sign wrong; there should be a line:
[MATH]0=k\log_2 10 - 2\hspace2ex[/MATH] (not +2) (and remembering that [MATH] \log_2 1=0[/MATH]).
so [MATH]k=\dfrac{2}{\log_2 10}[/MATH]
Answer: [MATH]c=-2, k=\dfrac{2}{\log_2 10}[/MATH]
(Incidentally, the original expression can simplify to [MATH]y=2(\log_{10} x - 1)[/MATH]).

just re wrote my working. I assume this is correct now? And would you think it is appropriate to find the decimal version of ‘k’, or just leave it.

 

[lex]

That's perfect. It is of course perfect to leave it like this.
However if in an exam I would also give a rounded decimal answer as well, because you have nothing to lose by putting it in and something to gain if for some strange reason, they have this in the mark scheme! You are protecting yourself from the marker/examiner.