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The question says: Given $\log_n(9)=x$ and $\log_n(4)=y$, find $\log_n(12)$. I'm just really stuck on how to work this out.

ndhanson3
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Lily
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  • This question uses the logarithm laws. Do you know how to find $\log_n(9\times 4)$ using that information? Or perhaps $\log_n(\frac{9}{4})$ or $\log_n(9^5)$? – ndhanson3 Mar 11 '21 at 00:22
  • I'm just confused as to what the value of n is and if I even need to work it out. I've only just started logarithms so it's very new to me – Lily Mar 11 '21 at 00:24
  • You do not need to find the value of $n$. The answer is the same for any value of $n$. – ndhanson3 Mar 11 '21 at 00:25

2 Answers2

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So we have given that $\log_{n} 9=x=x$ $\log_{n} 4=y$.

$$\log_{n} 9=\log_{n} 3^{2}=2\log_{n} 3=x$$ $$\log_{n} 3=\frac{x}{2}$$ $$\log_{n}12=\log_{n}3*4=\log_{n}3+\log_{n}4=\frac{x}{2}+y$$

So,

$$\log_{n}12=\frac{x}{2}+y$$

In this the answer doesen't change because of value of n. Just use normal logarithm laws. I just used,

  1. $\log_{n} (x^y) = y\log_{n} (x)$ = Log of a power
  2. $\log_n (xy) = \log_{n} (x)+\log_{n} (y)$ = Product rule.
Tharindu Gimras
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remember $log_n(ab) = log_n(a) + log_n(b) $

also $log_n(a^{c})= c\cdot log_n(a)$

Notice: $12 = 3 \cdot 4$

$$log_n(12) = log_n(4) + log_n(3)$$ We have $log_n(4)$ but not $log_n(3)$

But $3^{2} = 9$

So, $log_n(9) = log_n(3^{2}) = 2log_n(3) = x$ ,therefore $log_n(3) = \frac{x}{2}$

Therefore $$log_n(12) = log_n(4) + log_n(3) = y + \frac{x}{2}$$

spectralblue
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