# Logarithm question

• May 5th 2011, 11:59 AM
HighlyFlammable
Logarithm question
Hi there, I'm having difficulty with this question:

Two variables, x and y, are connected by the law y = a^x. The graph of log4y against x is a straight line passing through the origin and the point A(6,3). Find the value of a.

Thanks in advance :)
• May 5th 2011, 12:20 PM
TheEmptySet
Quote:

Originally Posted by HighlyFlammable
Hi there, I'm having difficulty with this question:

Two variables, x and y, are connected by the law y = a^x. The graph of log4y against x is a straight line passing through the origin and the point A(6,3). Find the value of a.

Thanks in advance :)

If we make a table of values you will see that

$\displaystyle \begin{array}{l|c|r} x & y & \log_{4}(y) \\ \hline 0 & 1 & 0 \\ 1 & a & log_{4}(a ) \\ 2 & a^2 & 2log_{4}(a ) \\ 3 & a^3 & 3log_{4}(a ) \\ 4 & a^4 & 4log_{4}(a ) \\ 5 & a^5 & 5log_{4}(a ) \\ 6 & a^6 & 6log_{4}(a ) \\ \end{array}$

Now we can pick the correct ordered pair to get

$\displaystyle 6log_{4}(a )=3 \iff log_{4}(a ) =\frac{1}{2} \iff a=4^{\frac{1}{2}}=2$
• May 5th 2011, 12:26 PM
Quacky
Quote:

Originally Posted by HighlyFlammable
Hi there, I'm having difficulty with this question:

Two variables, x and y, are connected by the law y = a^x. The graph of log4y against x is a straight line passing through the origin and the point A(6,3). Find the value of a.

Thanks in advance :)

The wording of the question confused me too.

My initial thoughts were this:

$\displaystyle Log_4(y)=x$
$\displaystyle y=4^x$
$\displaystyle a^x=4^x$
$\displaystyle a=4$

But this does not satisfy either of the given points.Therefore it must be wrong.

The reason is that the x's and y's in the question seemed slightly inconsistent with their meaning. I guess I need to brush up on my understanding of language!

I decided to start with the straight line passing through $\displaystyle (6,3)$ and the origin. The gradient of this line is $\displaystyle \frac{3}{6}=\frac{1}{2}$ The y-intercept is $\displaystyle 0$.

So the line they are referring to is $\displaystyle y=\frac{1}{2}x$.

Now, I assumed then that:

$\displaystyle Log_4{(y)}=\frac{1}{2}x$

...which, when you solve for $\displaystyle a$, gives the same answer as TES.

After rereading the question, I realised that it does make sense but that I was incapable of grasping its meaning.(Doh)