# First Year Maths Question (1 Viewer)

#### cagedbird17

##### New Member

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

##### New Member
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.

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#### cagedbird17

##### New Member
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

##### Retired '16
Drognoski might be able to help if he is on. He is really good with maths, I mean really really good.

#### cagedbird17

##### New Member
thanks, I'll try sending Drognoski a message

#### cagedbird17

##### New Member
Anyone got any ideas?

#### dan964

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

My guess would be y=3*10^x

#### dan964

##### MOD
Moderator
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

##### New Member

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

##### Well-Known Member
Yep, it's $\bg_white y=3\times 10^x$. 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 = 10d. 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 $\bg_white m=\frac{d_2 - d_1}{2-1}$, where $\bg_white d_2$ is the distance along the y-axis at x = 2 (which is log300, as y = 300 here) and $\bg_white d_1$ 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 $\bg_white 2-1$.)

So the slope of this straight line is $\bg_white \frac{\log_{10} 300 - \log_{10}30}{2-1}=1$ (using log laws to simplify).

So the equation of the straight line is $\bg_white \text{distance along } y\text{-axis} = 1\cdot x + \text{distance along }y\text{-axis when }x=0$ (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. $\bg_white \log_{10}y=x+\log_{10}3\Rightarrow y = 10^{x+\log_{10}3}=3\times 10^x$ (using log laws and index laws).

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#### cagedbird17

##### New Member
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.

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#### Drongoski

##### Well-Known Member
Just saw the question.

By inspection:

f(x) = 3 x 10x

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

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