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Firs time I've encountered this type of question where there is a number in front of the log, and it's throwing me a bit. I'm fine with using the logarithmic laws to simplify, but not sure if I need to get rid of those numbers first, and if so, how? thought perhaps the power law would apply but obviously there's no powers so I'm really not sure how to approach this one.

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Use that $$ a\log\left(b\right)=\log\left(b^a\right) $$ (and then :) $$\log\left(a\right)+\log\left(b\right)=\log\left(ab\right)$$

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HINT

There are 'hidden powers' in this expression; all the logarithms are of the form $a\log2^n$ where $a$ and $n$ are integers.

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$2log4=2log(2^2)=2\cdot 2log2=4log2$, using $log(2^2)=2log2$

$3log8=3log(2^3)=3\cdot 3log2=9log2$

$2log4+3log8-log2=4log2+9log2-log2=12log2$

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Considering $\log(x)=\log_{10}(x)$

I have just made this consideration because the expression is nicely solved for $\log_{2}(x)$ if it is perhaps a typo. For $\log_{2}(x)$ we would have

$$2\log_{2}(4)+3\log_{2}(8)-\log_{2}(2) = 2\cdot 2 + 3 \cdot 3 - 1 = 12$$


\begin{align} 2\log(4)+3\log(8)-\log(2) & = \log(4^2)+\log(8^3)-\log(2)\\ & = \log(4^2)+\log(8^3)-\log(2) \\ & = \log(16)+\log(512)-\log(2)\\ & = \log(8192)-\log(2) \\ & = \log \left(\frac{8192}{2}\right) \\ & = \log(4096)\\ & = \log(2^{12})\\ & = \boxed{12\log(2)\approx 3.6}\\ \end{align}

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Use that $$ a\log\left(x\right)=\log\left(x^a\right) $$ (and then :) $$ay+by+cy=(a+b+c)y \phantom{1} where \phantom{1} y=\log\left(x\right)$$

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