1. Express and calculate...

Hi,

If $\displaystyle \log_9 x = a$ and $\displaystyle \log_3 y = b$, express $\displaystyle xy$ and
$\displaystyle \frac{x}{y}$ as powers of $\displaystyle 3$.
If $\displaystyle xy = 243$
and $\displaystyle \frac{x}{y} = 3$, calculate $\displaystyle a$ and $\displaystyle b$.

I got the expression part but not the calculation:

$\displaystyle \log_9 x = a,\,\,\,\,\,\,\,\,\,\,\log_3 y = b$

$\displaystyle x = 9^a = 3^{2a}\,\,\,\,\,\,\,\,y = 3^b$

$\displaystyle xy = 3^{2a} \times 3^b$
$\displaystyle xy = 3^{2a + b}$

$\displaystyle \frac{x}{y} = \frac{3^{2a}}{3^b}$
$\displaystyle = 3^{2a - b}$

2. What don't you get? It looks fine.

3. $\displaystyle xy=243=3^5$

$\displaystyle xy=3^{2a+b}$

$\displaystyle \therefore 3^5=3^{2a+b}$...............(I)

likewise, $\displaystyle \frac{x}{y}=3 \;and\;\frac{x}{y}=3^{2a-b}$

$\displaystyle \therefore 3^{1}=3^{2a-b}$................(II)

you can find a and b from equations (I) and (II)

4. Originally Posted by harish21
$\displaystyle xy=243=3^5$

$\displaystyle xy=3^{2a+b}$

$\displaystyle \therefore 3^5=3^{2a+b}$...............(I)

likewise, $\displaystyle \frac{x}{y}=3 \;and\;\frac{x}{y}=3^{2a-b}$

$\displaystyle \therefore 3^{1}=3^{2a-b}$................(II)

you can find a and b from equations (I) and (II)
Ah, I made an attempt like that but I had no faith in it. Thanks.

I'm not sure if this works:

$\displaystyle 2a + b = 5$
$\displaystyle 2a - b = 1$

$\displaystyle 4a = 6$
$\displaystyle a = \frac{3}{2}$

$\displaystyle 2(\frac{3}{2}) - b = 1$
$\displaystyle 3 - b = 1$
$\displaystyle b = 2$

5. -Well Done!

You can check whether your values for a and b are correct or not by plugging their values back in equations (I) and (II)

Check:

Equation (II)

$\displaystyle 3^1=3^{2a-b}$

$\displaystyle 3=3^{2\cdot\frac{3}{2}-2}$

$\displaystyle 3=3$