# Help with two log proof simultaneous questions please

• Dec 27th 2011, 10:23 AM
[SOLVED] Help with two log proof simultaneous questions please
Hi,

My elderly father has been stumped by the following questions for the past few days. It's beyond my knowledge. I wondered if anyone here might be able to solve these two attached questions 10 and 16 (I did try to write them out and use LATEX code but to no avail)?

Here's my efforts using LATEX
Q10: Show that log_{16} $\displaystyle (xy)$= $\displaystyle \frac{1}{2}$ $\displaystyle log_4$ $\displaystyle x$+$\displaystyle \frac{1}{2}$ $\displaystyle log_4$ $\displaystyle y$. Hence, or otherwise, solve the simutaneous equations

log_16 $\displaystyle (xy)$=3 $\displaystyle \frac{1}{2}$

(($\displaystyle log_4$ $\displaystyle x$) / ($\displaystyle log_4$ $\displaystyle y$)) = -8

Q16: Express log_9 $\displaystyle x$$\displaystyle y$ in terms of $\displaystyle log_3$ $\displaystyle x$ and $\displaystyle log_3$ $\displaystyle y$.
Without using tables, solve for x and y the simultaneous equations

log_9 $\displaystyle x$ $\displaystyle y$= $\displaystyle \frac{5}{2}$
$\displaystyle log_3$ $\displaystyle x$ $\displaystyle log_3$ $\displaystyle y$ = -6
• Dec 27th 2011, 10:53 AM
Siron
Re: Help with two log proof simultaneous questions please
To show that: $\displaystyle \log_{16}(xy)=\frac{1}{2}\log_4(x)+\frac{1}{2}\log _4(y)$

Rewrite:
$\displaystyle \frac{1}{2}\log_4(x)=\log_4(\sqrt{x})=\frac{\log_{ 16}(\ldots)}{\log_{16}(\ldots)}$

Complete the $\displaystyle \ldots$ and do exactly the same for the other term.

Try to do something similar for the other question Q16.

For the simultaneous equations: you have given two equations and two variables $\displaystyle x$ and $\displaystyle y$ so use the substitution method (do you know how this works?)
• Dec 27th 2011, 12:57 PM
Re: Help with two log proof simultaneous questions please
Hi Siron,

Many thanks for your speedy response. I've spoken to my Dad and he does understand what you are saying. He got as far as the following, however, its the Substitution method that he says doesn't seem to be giving him the result he expects.

\frac{\log_4(x)}{\log_4(x)} = -8

therefore \log_4(x) - \log_4(y) = -8

\log_16(xy) = \frac{7}{2}
therefore \frac{1}{2}\log_4(xy) = \frac{7}{2}
therefore log_4(xy) = 7

Can you please point out where he is going wrong? He knows that x = 4^(-1) = \frac{1}{4}, and he knows that y= 4^(8). He just can't work out how.

On a personal note; I have downloaded the LATEXT how-to PDF but my previews have errors or look like the above. If you could tell me how you type LATEXT above correctly I would appreciate that.

Thank you kindly for your help once again.
• Dec 27th 2011, 01:06 PM
Plato
Re: Help with two log proof simultaneous questions please
Actually $\displaystyle \frac{\log_4(x)}{\log_4(y)}=-8$ gives $\displaystyle \log_4(x)=-8\log_4(y)$ or $\displaystyle x=y^{-8}.$
• Dec 27th 2011, 01:15 PM
Siron
Re: Help with two log proof simultaneous questions please
To work with LaTEX you have to use the brackets: [ tex ] ... [ /tex ].
We are given the simultaneous equations:
$\displaystyle \left \{ \begin{array}{ll} \log_{16}(xy)=\frac{7}{2} \\ \frac{\log_4(x)}{\log_4(y)}=-8 \end{array} \right .$

We can rewrite the first equation with the given:
$\displaystyle \left \{ \begin{array}{ll} \frac{1}{2}\left[\log_4(x)+\log_4(y)\right]=\frac{7}{2} \\ \frac{\log_4(x)}{\log_4(y)}=-8 \end{array} \right .$

Solving the system:
$\displaystyle \left \{ \begin{array}{ll} \frac{1}{2}\left[\log_4(x)+\log_4(y)\right]=\frac{7}{2} \\ \log_4(x)=-8\log_4(y) \end{array} \right .$

By substituting the second equation in the first we obtain:
$\displaystyle \left \{ \begin{array}{ll} \frac{1}{2}\left[-8\log_4(y)+\log_4(y)\right]=\frac{7}{2} \\ \log_4(x)=-8\log_4(y) \end{array} \right .$

$\displaystyle \Leftrightarrow \ldots$

I guess you have switched the answers, beside my calculations I obtain: $\displaystyle x=4^8$ and $\displaystyle y=\frac{1}{4}$ which is correct by substituting them into the system.

Proceed.

The other system of equations is similar to solve.
• Dec 28th 2011, 12:50 AM