Original post by Nithu05
The graph of log10y against x is a straight line
Then it shows a point of (4,6) and a y intercept of 2.
i) find the equation for log10y in terms of x
ii) find the equation for y in terms of x
Then it shows a point of (4,6) and a y intercept of 2.
i) find the equation for log10y in terms of x
ii) find the equation for y in terms of x
I got log10y=x+2. Is this right?
Have you tried putting the point (4,6) into your equation?
Original post by qweqworiet
Have you tried putting the point (4,6) into your equation?
I worked out the gradient using the points (4,6) and (0,2). However, the mark scheme says the gradient is 3. Could you please explain?
Original post by Nithu05
The graph of log10y against x is a straight line
Then it shows a point of (4,6) and a y intercept of 2.
i) find the equation for log10y in terms of x
ii) find the equation for y in terms of x
Then it shows a point of (4,6) and a y intercept of 2.
i) find the equation for log10y in terms of x
ii) find the equation for y in terms of x
Can you upload a picture of the original question?
(edited 1 year ago)
Original post by John mburu
I can help with the step by step calculation on how to go about it.
Ok couldn't you just post the answer here?
Original post by Skiwi
Can you upload a picture of the original question?
Here’s the picture
Original post by John mburu
okay let me work on it in my book and the I will post it.
Thank you so much
Original post by Nithu05
Thank you so much
I agree with your answer.
No charge.
Original post by mqb2766
I agree with your answer.
No charge.
No charge.
So is the mark scheme wrong?
Original post by Nithu05
So is the mark scheme wrong?
Would appear so.
log10(y) = 6
means y = 10^6, whcih is not 10^(3*4+2).
(edited 1 year ago)
Original post by mqb2766
Would appear so.
Thanks so much. Could you please check my other post on a vector question?
Original post by Nithu05
Thanks so much. Could you please check my other post on a vector question?
Ok, I edited the previous post slightly as you were replying.
Original post by Nithu05
Thanks so much. Could you please check my other post on a vector question?
https://www.thestudentroom.co.uk/showthread.php?t=7228005&p=97314805#post97314805
Original post by John mburu 
follow me on Instagram for more @George_ruben12
follow me on Instagram for more @George_ruben12
Thank you! Although, we haven't learnt antilogs yet.
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