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LynnM · Aug 31, 2010, 02:35 PM

So, here's another question that I am having trouble with:

[(2log(5)x) / (log(5)3)] - log(5)y = 2

I followed the steps that were used in a sample question just like this one and I still got the wrong answer... or I just don't know how to finish it. Here's what I've done:

(2/log(5)3) * log(5)x = 2 + log(5)y

log(5)x^{2/log(5)3} = 2log(5)5 + log(5)y

log(5)x^{2/log(5)3} = log(5)(25y)

x^{2/log(5)3} = 25y

y = (x^{2/log(5)3}) / 25

The text solution is:

y = [(x^{2})/9] ^{1/(log(5)3)}

I'm lost. Help please!

Unknown008 · Aug 31, 2010, 10:22 PM

Ok, after a few tries, this is what I did:

$\frac{2log_5 x}{log_5 3} - log_5 y = 2$

I used the fact that: $\frac{log_c a}{log_c b} = log_b a$

$2log_3 x - log_5 y = 2$

$log_3 x^2 -2 = log_5 y$

And now, we know that $2 = 2log_3 3 = log_3 9$

$log_3 x^2 -log_3 9 = log_5 y$

$log_3 $\frac{x^2}{9}$= log_5 y$

I now convert back to base 5.

$\frac{log_5 $\frac{x^2}{9}$}{log_5 3}= log_5 y$

$log_5 $\frac{x^2}{9}$= (log_5 3)log_5 y$

$log_5 $\frac{x^2}{9}$= log_5 y^{log_5 3}$

$\frac{x^2}{9}= y^{log_5 3}$

$$\frac{x^2}{9}$^{\frac{1}{log_5 3}}= y$

LynnM · Sep 1, 2010, 07:59 AM

Tricky, tricky! Thanks! Is [(log(c)a) / log(c)b)] = log(b)a commonly used? My text never provided this. So, was my initial answer incorrect even though similar questions were answered that way?

Unknown008 · Sep 1, 2010, 08:06 AM

I didn't see anything wrong with your answer. If we could turn it around, I'm sure that we'd get the answer by the book sooner or later.

I use this identity often, especially to calculate values. For example:

$log_5 35 = \frac{log_{10} 35}{log_{10} 5}$

Since calculators cannot find the log to all the bases but only 10, you have to divide the log of the number by the log of the base.

Of course, now you could also use $ln$ which is log to the base of 'e'.

LynnM · Sep 1, 2010, 08:14 AM

Hmmm, not my cup of tea, that's for sure! Thanks for all of your help. I'm sure I will have more questions posted in the near future.

Also, sorry about the mutiple threads. I thought I forgot to submit my first. It's not possible to remove it, is it?

Unknown008 · Sep 1, 2010, 08:22 AM

Not now... but we learn by mistakes, right? :)