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Logarithmic Functions - argh - help please


Charlie

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I've been trying hours to do 2 questions for my maths homework. I've got the answer but don't have a clue how to get there. Could someone point me in the right direction please?

 

1) The sketch below shows the graph y=log10 (x+c). Find the value of c.

 

graph13so.jpg

 

2) The sketch below shows the graph y=alog2 (x+b). Find a and b

 

graph24hv.jpg

 

I've worked out b to be 4 (which is right), I just can't get a.

 

 

Answers:

1) 3

2) a= 3 1/3 b= 4

 

 

------

 

Thanks a lot in advance.

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i'm guessing this is the scottish equivalent to A Level work?

 

we do nothing like this at GCSE... though i'm gonna be doing it at A Level ^_^

 

GCSE = Standard Grade (what I did last year)

 

So if A level is the one after GCSE then that is what it is. Highers, they're called. And they're bloody hard.

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Bugger knows.....we only did log to the base e.

 

 

Wait for Odwin methinks. Althoughhe did say he hated pure maths. There was a dude in the Uni thread who said he was doing maths at Uni. Forget his name though.

 

I hope he drops by in the next 30 mins. This is meant to be in for tomorrow....

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Dude, if you can't do it, then just tell he teacher you can't do it. They can't berate you for something you can't do. Especially if you've tried for two hours.

 

Yeah, I wil. At least I've actually tried the question. She just said "there better not be any blanks".

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I think the log10 means it is base 10 so we have:

y = log10 (x+c)

0 = log10 (c -2) [from graph]

10^0 = c-2

1 = c-2 [anything to power of 0 is 1]

1+2 = c

therefore c = 3

 

log2 is log to base 2, so solve simultaneously:

1) 10 = alog2 (4+b)

2) 0 = alog2 (b-3)

we'll work with 2) first:

0/a = log2 (b-3)

0 = log2 (b-3)

2^0 = b-3

1 = b-3

1+3 = b

therefore b = 4

plug b = 4 into 1):

10 = alog2 (4+4)

10 = alog2 (8)

10 = log2 (8^a) [one of the rules, look in your txt bk]

2^10 = 8^a

1024 = 8^a

ln1024/ln8 = 10/3 = 3 1/3

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I think the log10 means it is base 10 so we have:

y = log10 (x+c)

0 = log10 (c -2) [from graph]

10^0 = c-2

1 = c-2 [anything to power of 0 is 1]

1+2 = c

therefore c = 3

 

log2 is log to base 2, so solve simultaneously:

1) 10 = alog2 (4+b)

2) 0 = alog2 (b-3)

we'll work with 2) first:

0/a = log2 (b-3)

0 = log2 (b-3)

2^0 = b-3

1 = b-3

1+3 = b

therefore b = 4

plug b = 4 into 1):

10 = alog2 (4+4)

10 = alog2 (8)

10 = log2 (8^a) [one of the rules, look in your txt bk]

2^10 = 8^a

1024 = 8^a

ln1024/ln8 = 10/3 = 3 1/3

 

Thanks. My teacher showed me how to do it today and her way was the same. :)

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