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Full Question is:

Given that $\log_{24} 12 =0.782$, find the value of $\log_{24} 2$ .

How do I / should I set this out as well, formally?

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    Please post correct question. – 2017-01-16
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    Hint: What is $\log_{24} 24$? – 2017-01-16

2 Answers 2

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$$\log_{24}12=\log_{24}\left(\frac{24}{2}\right)=\log_{24}24-\log_{24}2=1-\log_{24}2$$

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    By "hint" you mean "almost complete solution" ;-P – 2017-01-16
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    Even easier, start with $\log_{24}2=\log_{24}\frac{24}{12}$. – 2017-01-16
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    @DavidK: I think they are equivalent. The idea and the calculation. But in fact, it is another good point. – 2017-01-16
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    Oh, yes, it is exactly the same idea and steps, except that you don't need to follow it with "solve for $x$ in $0.782 = 1 - x$." – 2017-01-16
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    @Cicada: is it clear? – 2017-01-18
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$12=\dfrac{24}{2}$

$\log_{24} 12=\log_{24}{\frac{24}{2}}=\log_{24} 24 - \log_{24} 2$

$1-\log_{24}{2}= 0.782$

$\log_{24}{2} = 0.218$

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    Correct your calculation. – 2017-01-16
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    @KanwaljitSingh I can't say anything more than thanks ! – 2017-01-16
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    You can now say thanks. Your welcome bro :-) – 2017-01-16
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    Note that if you start by observing that $2=\frac{24}{12}$ and proceed from there, you quickly get $\log_{24}2$ isolated on one side of an equation without the risk of errors while trying to solve $1-\log_{24}{2}= 0.782$. – 2017-01-16