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I am having problems working out how to approach solving this problem :

$$\left(\frac{81}{16}\right)^n = \frac{32}{243}$$

How do I go about working out $^n$? Please step by step if possible.

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up vote 7 down vote accepted

Using laws of exponents, you can rewrite the equation as follows.

$$\left(\frac{81}{16}\right)^n = \frac{32}{243}$$

$$\left(\frac{3^4}{2^4}\right)^n = \frac{2^5}{3^5}$$

$$\left(\frac{3}{2}\right)^{4n} = \left(\frac{2}{3}\right)^5 $$

$$\left(\frac{2}{3}\right)^{-4n} = \left(\frac{2}{3}\right)^5$$

From this, it follows that

$$-4n = 5$$

$$n = -\frac{5}{4}$$

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Very clear thank you! – لَا إِلٰهَ إِلَّا الله Mar 16 '13 at 12:39

Here is an approach.

$\displaystyle \left(\frac{81}{16}\right)^n = \frac{32}{243}$

$\displaystyle \left(\frac{3^4}{2^4}\right)^n = \left(\frac{2^5}{3^5}\right)$

$\displaystyle \left(\frac{2}{3}\right)^{-4n} = \left(\frac{2}{3}\right)^5$

So, $-4n = 5 \rightarrow n = -\frac{5}{4}$

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I'm not all that sure I'd say this method is "without logs". You're finding that the logarithm to the base $2/3$ of $32/243$ is $5$, and the logarithm of $(81/16)^n$ is $-4n$. – Michael Hardy Mar 16 '13 at 12:49
@MichaelHardy: Yeah - after thinking about it, I am going to delete that statement - thanks. – Amzoti Mar 16 '13 at 13:04
Nice work here! – amWhy Apr 20 '13 at 0:14

apply $\log$ on both side and reduce the equation

$$n \log\left(\frac{81}{16}\right) = \log\left(\frac{32}{243}\right) $$ $$ n(\log 81 -\log 16) = \log 32 - \log 243$$ $$ n = \frac{\log 32 - \log 243}{\log 81 -\log 16} $$

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After that you can say $\log32-\log243$ $=\log(2^5)-\log(3^5)$ $=5\log2-5\log_3$ $=5\log(2/3)$. And $\log81-\log16$ $=\log(3^4)-\log(2^4)=4(\log3-\log2)$ $=-4\log(2/3)$. So the fraction simplifies by canceling $\log(2/3)$ and you get $5/(-4)$. – Michael Hardy Mar 16 '13 at 12:47

The point is $3^4=81$, $2^4=16$, $3^5=243$, and $2^5=32$:

$$\left(\frac{81}{16}\right)^n = \frac{32}{243}$$ $$ \left(\frac 3 2\right)^{4n} = \left(\frac 3 2 \right)^{-5} $$ So $4n=-5$.

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