# Thread: determine an expression for log3 interms of x and y

1. ## determine an expression for log3 interms of x and y

no cal multiple choice
Given Log 6 = x Log 8 = y determine an expression for log3 interms of x and y.

a: x/3 - y
b: x-y/3
c: y/3-x
d y-x/3

Don't understand how to turn log 3 into log 6 and log8.. help

ThankYou

2. remember log(xy) = log(x) + log(y). So
log 6 = log (2*3) = log 2 + log 3.
Also log 8 = log 2*2*2 = log 2 + log 2 + log 2 = 3log2.
Can you proceed?

3. Sorry.. i can't proceed.
If log2 + log3
log2 + x

3log2 = log2^3

..ah..>.<

$\displaystyle \log{6} = x\log{8}$

$\displaystyle \log{(2\cdot 3)} = x\log{\left(2^3\right)}$

$\displaystyle \log{2} + \log{3} = 3x\log{2}$

$\displaystyle \log{3} = 3x\log{2} - \log{2}$.

Call this Equation 1.

Also, remember that

$\displaystyle \log{6} = y$

$\displaystyle \log{(2\cdot 3)} = y$

$\displaystyle \log{2} + \log{3} = y$

$\displaystyle \log{2} = y - \log{3}$.

Substitute this back into Equation 1.

$\displaystyle \log{3} = 3x(y - \log{3}) - (y - \log{3})$

$\displaystyle \log{3} = 3xy - 3x\log{3} - y + \log{3}$

$\displaystyle 0 = 3xy - 3x\log{3} - y$

$\displaystyle 3x\log{3} = 3xy - y$

$\displaystyle \log{3} = y - \frac{y}{3x}$.

So none of your possibilities are correct...

5. Hello, hovermet!

Given: .$\displaystyle \log 6 = x,\;\log 8 = y$
Determine an expression for $\displaystyle \log3$ in terms of $\displaystyle x$ and $\displaystyle y$

. . $\displaystyle (a)\;\;\frac{x}{3} - y \qquad (b)\;\;x-\frac{y}{3} \qquad (c)\;\;\frac{y}{3}-x \qquad (d)\;\;y-\frac{x}{3}$

Note that: .$\displaystyle \log 8 \,=\,y \quad\Rightarrow\quad \log(2^3) \,=\,y \quad\Rightarrow\quad 3\log 2 \,=\,y \quad\Rightarrow\quad \log 2 \,=\,\frac{y}{3}$

So we have: .$\displaystyle \log 3 \;\;=\;\;\log\left(\frac{6}{2}\right) \;\;=\;\;\log 6 - \log 2 \;\;=\;\;x - \frac{y}{3}$

6. Thanks guys.. it seems like i forgot an AND between log 6 = x Log 8 = y ^^;
In fact, i done the log test pretty well today with the understanding of the mistaken question Thanks^^