If $60^a = 3$ and $60^b = 5$, then find $12^{\frac{1-a-b}{2(1-b)}}$

Question

If $60^a = 3$ and $60^b = 5$, then find $12^{\frac{1-a-b}{2(1-b)}}.$

I have included an answer of my own in case you're stuck.

Answer 1

as $$a=\log _{ 60 }{ 3 } \\ b=\log _{ 60 }{ 5 } $$ we have $$12^{ \frac { 1-a-b }{ 2(1-b) } }=12^{ \frac { 1-\log _{ 60 }{ 3 } -\log _{ 60 }{ 5 } }{ 2(1-\log _{ 60 }{ 5 } ) } }=12^{ \frac { 1-\left( \log _{ 60 }{ 3 } +\log _{ 60 }{ 5 } \right) }{ 2(1-\log _{ 60 }{ 5 } ) } }=12^{ \frac { \log _{ 60 }{ 60 } -\left( \log _{ 60 }{ 15 } \right) }{ 2(\log _{ 60 }{ 60 } -\log _{ 60 }{ 5 } ) } }=\\ =12^{ \frac { \log _{ 60 }{ \frac { 60 }{ 15 } } }{ 2\log _{ 60 }{ \frac { 60 }{ 5 } } } }=12^{ \frac { \log _{ 60 }{ 4 } }{ \log _{ 60 }{ 144 } } }={ 12 }^{ \log _{ 144 }{ 4 } }={ 4 }^{ \frac { 1 }{ 2 } \log _{ 12 }{ 12 } }=2\\ $$

Answer 2

Here's my way of doing this:

This problem is easier if we turn the first two equations into logs: $a = \log_{60} 3$, $b = \log_{60} 5$. Then

\begin{align*} 12^{ \frac{1-a-b}{2(1-b)} } &= 12^{ \frac{1 - \log_{60} 3 - \log_{60} 5}{2(1 - \log_{60} 5)} } \end{align*}

Using the fact that $1 = \log_{60} 60$, we can combine the linear combination of logs into one log. In fact, adding and subtracting logs in this way is how multiplication and division is done with log tables and slide rules.

\begin{align*} 12^{ \frac{1-a-b}{2(1-b)} } &= 12^{ \frac{\log_{60} 60/(3 \cdot 5)}{2(\log_{60} 60/5)} } \\ &= 12^{ \frac{\log_{60} 4}{2\log_{60} 12} } \\ &= 12^{\frac{1}{2} \log_{12} 4} \\ &= 4^{1/2} \\ &= \boxed{2}. \end{align*}

Answer 3

Notice $\ \ \begin{align}&ABC^{\large a} = A\ \Rightarrow\ a = \ell A\\ &ABC^{\large b} = B\ \Rightarrow\ b = \ell B\end{align}\ \ $ where $\,\ \ \ell N\, :=\, \log_{ABC}\! N$

So if $\,\ x = AC^{\Large \frac{(1-a-b)}{2(1-b)\ \,}}\ $ then taking $\,\ell\,$ of it and using above to replace $\,a,b\,$ yields

$\quad\ \ \ell x = \dfrac{\ell ABC-\ell A-\ell B}{2\,(\ell ABC-\ell B)}\,\ell AC\,=\,\dfrac{\ell C}{2\,\ell AC}\,\ell AC\, =\, \dfrac{\ell C}{2}\,\Rightarrow\ x = C^{1/2}$

Yours is the special case $\ A,B,C = 3,5,4\ $ so $\ x = C^{1/2} = 4^{1/2} = 2$

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Remark $\ $ From above we see that the identity arises essentially as follows

$${\large C}\, =\, {\large AC}^{\,\dfrac{\ell C}{\ell AC}} =\, {\large AC}^{\dfrac{\ell(ABC/AB)}{\ell(ABC/B)}}\!\! =\, {\large AC}^{\dfrac{1-\ell A-\ell B}{1 - \ell B}}\!\! =\, {\large AC}^{\dfrac{1-a-b}{1 - b}}$$