First Year Maths Question (2 Viewers)

[cagedbird17]

I'm not sure if this is the right place to post this thread, but could someone please help me with this question, I don't even know where to start.

 

[cagedbird17]

here's how I tried to approach the question

since the log scale is one the y axis

logy= mx+ b

y= 10^ (mx + b)

y= 10^(mx) x10^b

I think this should be the formula, but then again i have no clue what I'm doing. I tried to say that b= 3 but i just don't know where i am going with this.

someone help please!

 

[cagedbird17]

yeah- I'm still stuck on this question.

I also tried this method

y= A x 10^kx

find A when x= 0 ( y= 3 at x=0)

A=3

y= 3x 10^kx

i tried to substitute in the point (1, 30) to find K but i kept getting 3.0959 which doesn't make sense.

anyone's thoughts on this would be greatly appreciated.

 

[Speed6]

Drognoski might be able to help if he is on. He is really good with maths, I mean really really good.

 

[cagedbird17]

thanks, I'll try sending Drognoski a message

 

[cagedbird17]

Anyone got any ideas?

 

[dan964]

first just consider that
f(0)=3
f(1)=30
f(2)=300

My guess would be y=3*10^x

 

[dan964]

yeah- I'm still stuck on this question.

I also tried this method

y= A x 10^kx

find A when x= 0 ( y= 3 at x=0)

A=3

y= 3x 10^kx

i tried to substitute in the point (1, 30) to find K but i kept getting 3.0959 which doesn't make sense.

anyone's thoughts on this would be greatly appreciated.

y=3*10^kx
Sub (1, 30) in

30=3*10^k
Divide both sides by 3
10=10^k
Take ln of both sides
gives k=1

 

[cagedbird17]

Thank you for your help!

I cursed myself when I realized that I should have divided by three before taking the log of everything.

once again- thank you so much.

 

[InteGrand]

Yep, it's . Here's a way to derive it:

The "y-axis", corresponds to the following: going up a physical distance of d units on the y-axis represents the value y = 10^d. So d = log₁₀ y, where y is the reading on the y-axis when you go up d units.

The slope of the line is , where is the distance along the y-axis at x = 2 (which is log₃₀₀, as y = 300 here) and is the distance along the y-axis at x = 1 (which is log₁₀ 30, as y = 30 here). (Note that the x-axis is linear and 1:1 scale, i.e. a distance of d along it represents a value of d for x, which is why the denominator in the slope formula is just .)

So the slope of this straight line is (using log laws to simplify).

So the equation of the straight line is (this is just slope-intercept form of a straight line when the axes are to scale, i.e. a distance of d along each axis represents a value of d of that variable).

i.e. (using log laws and index laws).

 

[cagedbird17]

I think I understand the process a bit better now, thanks InteGrand. Before I was just using a formula that I knew applied to the question but I didn't exactly know what was going on.

 

[Drongoski]

Just saw the question.

By inspection:

f(x) = 3 x 10^x

as provided, thru other approaches, by dan and InteGrand.