# determine an expression for log3 interms of x and y

• Dec 7th 2009, 08:44 PM
hovermet
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
• Dec 7th 2009, 08:57 PM
qmech
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?
• Dec 7th 2009, 09:27 PM
hovermet
Sorry.. i can't proceed.
If log2 + log3
log2 + x

3log2 = log2^3

..ah..>.<
• Dec 7th 2009, 09:50 PM
Prove It

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

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

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

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

Call this Equation 1.

Also, remember that

$\log{6} = y$

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

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

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

Substitute this back into Equation 1.

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

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

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

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

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

So none of your possibilities are correct...
• Dec 8th 2009, 04:20 AM
Soroban
Hello, hovermet!

Quote:

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

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

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

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

• Dec 8th 2009, 01:38 PM
hovermet
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^^