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
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
Start with the fact that
$\displaystyle \log{6} = x\log{8}$
$\displaystyle \log{(2\cdot 3)} = x\log{\left(2^3\right)}$
$\displaystyle \log{2} + \log{3} = 3x\log{2}$
$\displaystyle \log{3} = 3x\log{2} - \log{2}$.
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...
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: .$\displaystyle \log 3 \;\;=\;\;\log\left(\frac{6}{2}\right) \;\;=\;\;\log 6 - \log 2 \;\;=\;\;x - \frac{y}{3}$