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Given is $$\log_6(a)=6$$

Simplify - $$\log_6 (1/a^7)$$ and the answer is supposed to be $-42$.

I don't understand what I'm supposed to do here?

user
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user591195
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  • There are identities about the logarithms of powers you have probably been taught that will help you here. – Ethan Bolker Nov 12 '18 at 00:13
  • The answer would be exactly the same if you are given $\log_k a=6$ and are asked to compute $\log_k(1/a^7)$. The value of $k$ is irrelevant. It could be $6$ or $e$ or $1/2$. – egreg Nov 12 '18 at 00:43

2 Answers2

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HINT

Recall that by definition

$$\log_6a=6 \iff 6^6=a$$

then

$$\log_6\left(\frac1{a^7}\right)=x \iff 6^x=\frac1{a^7}$$

or use that

  • $\log x^n=n\log x$
  • $\log \frac1x=-\log x$
user
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  • So do I replace a with 46656 and rise it to the 7th power? – user591195 Nov 12 '18 at 00:36
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    We have that $6^6=a\implies 6^{42}=a^7$. Can you conclude from here? – user Nov 12 '18 at 00:37
  • Ohhh yes thank you! – user591195 Nov 12 '18 at 00:42
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    There is no need to go to the exponential; using $\log_6(a^{-7})=-7\log_6a$ is the best path. – egreg Nov 12 '18 at 00:45
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    That’s a way by the definition the other one is use the formula for logarithm which of course are deduced from the same definition, that is $$\log_6(1/a^7)=\log_6 (a^{-7})=-7\log_6 a=-42$$ but the application by the definition is much more useful if you need to get confident with that. – user Nov 12 '18 at 00:48
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    @egreg I’ve just explained my aim here above. If one asker can’t solve a trivial problem like that I think itbis not very useful to give a solution by some “magic” formula he/she needs to memorize. As first choice I highly prefer a solution based on the definition. Anyway I’ve also indicated the formulas you are referring to for the solution. – user Nov 12 '18 at 00:53
  • @gimusi That's not a “magic formula”, but an important consequence of the main property of the logarithm. Everybody ought to know that $\log(x^y)=y\log x$. – egreg Nov 12 '18 at 07:51
  • @egreg Yes of course, I’m referring to the asker’s perspective when I call it “magic”. I’m not claiming that the formula is not important (I’ve also indicated that) but I think that it is more important to understand the properties from the basic definition insted to use a given formula. From there indeed one aquires the knowledge to prove those formulas and use that in a proper way. – user Nov 12 '18 at 08:00
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For any logarithm, $\log a^b=b \log a$, so in your case $$\log_6\frac 1{a^7}=-7\log_6a=-42$$ We don't have to solve for $a$ at all, though we could use $\log_6a=6$ to say $a=6^6$

Ross Millikan
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