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I am to evaluate $6\log_8(4)$ without a calculator. The answer is provided as 4 but I cannot see how to arrive at 4.

Ignoring the 6 at the beginning of the expression, $log_8(4)$ can be written as $8^x=4$.

Without the use of a calculator I cannot simplify that part any further. I know that the root or 3rd root of 8 is not 4.

I'm not sure if the leading 6 helps or if I am to multiply the end result by 6 or some part of the expression by 6 while doing the working out part?

Completely stuck here. How can I arrive at 4? Granular, baby steps appreciated.

[EDIT] I'd like to add that this is the textbook chapter I am working of. It's the absolute beginning of learning about logarithms. Looking at the comments and answer so far, there's reference to dividing logs which has not been covered in this book so far. Given the content of this chapter, I wonder if it's expected of me that I know how to solve this question?

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    $\begingroup$ Hint: What's $\log_82$? $\endgroup$
    – lulu
    Nov 18 '20 at 13:22
  • $\begingroup$ Is it 1/3? OK? Note I'm very new to logs, started reading about them last night. $\endgroup$
    – Doug Fir
    Nov 18 '20 at 13:25
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    $\begingroup$ Yes! $2^3=8\implies 8^{1/3}=2$. So, then, what is $\log_2 4$? $\endgroup$
    – lulu
    Nov 18 '20 at 13:30
  • $\begingroup$ Alright, I'm following this and it's making sense... it's 2? Still not seeing the picture holistically though? $\endgroup$
    – Doug Fir
    Nov 18 '20 at 13:32
  • $\begingroup$ No...$4=2^2$ so $\log_84=\log_8 2^2=2\log_82$. $\endgroup$
    – lulu
    Nov 18 '20 at 13:39
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Let's go step by step.

We start with the fact that $8=2^3$. That's equivalent to $2=8^{1/3}$ from which we immediately deduce that $$\log _{8} 2=\frac{1}{3}$$

Now, of course, we really wanted $\log_84$. But, as $4=2^2$ we have $$\log_84=\log_82^2=2\log_82=\frac 23$$ It follows that $$\boxed{6\log_84=6\times \frac 23=4}$$

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Using $\log_ac=\dfrac{\log_bc}{\log_ba}$ when all the logarithms remain defined,

$$\log_84=\dfrac{\log_24}{\log_28}=\dfrac{\log_2(2^2)}{\log_2(2^3)}=?$$

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Or you can see that $\log(a/b)=\log a -\log b$. So $\log_8 4=\log_8{(8/2})=\log_88-\log_8 2=1-\frac 1 3=\frac 2 3$

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$6\log_8(4)=\log_8(4^6)=\log_8((2^2)^6)=\log_8(2^{12})=\log_8((2^3)^4)=\log_8(8^4)=4$

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  • $\begingroup$ It would be still clearer, without any more computations, to write $\log_{8}(2^{12})=\log_{2^3}(2^{12})$. $\endgroup$
    – Bernard
    Nov 18 '20 at 13:28
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    $\begingroup$ I agree, but I wanted to avoid anything that would remotely look like the quotient formula for logarithms. (This is why it is a bit more verbose.) I could've also skipped all the steps except first and the last and just state that $4^6=8^4=4096$ but that would look like magic. Not sure - I tailored this answer to my own personal standard about clarity, YMMV. $\endgroup$ Nov 18 '20 at 13:34
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To deal with $8^x=4$, note $8=2^3$ and $4=2^2$.

So that $$ 8^x = 4 $$

$$ \Rightarrow (2^3)^x = 2^2 $$

$$ \Rightarrow 2^{3x} = 2^2 $$

$$ \Rightarrow 3x = 2 $$

$$ \Rightarrow x = \dfrac{2}{3} $$

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$$6\log_84=6\log_{2^3}{2^2}=6\times\frac{2}{3}\times\log_22=4\times 1=4$$

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  • $\begingroup$ What's this superscript notation for the argument of log? I've never seen it before. Any published references? $\endgroup$
    – Unit
    Nov 18 '20 at 14:00
  • $\begingroup$ No, I'm asking about the notation. Who writes "$\log_b^x$" instead of "$\log_b x$"? $\endgroup$
    – Unit
    Nov 19 '20 at 0:14
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Let, $6 log_8 (4) = x\\ \text{or, } log_8(4) = \frac{x}{6} \\ \text{or, } 4 = 8^\frac{x}{6} \\ \text{or, } 2^2 = 2^{3 × \frac{x}{6}} \\ \text{or, } 2 = \frac{x}{2} \because \text{equating powers of equal base} \\ \therefore x = 4$

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