## Thursday, April 23, 2015

### Find the value $12^\frac{1-(x+y)}{2(1-y)}$ given $3=60^x$ and $5=60^y$.

Without using the calculator and the help from logarithm, evaluate $\large12^\frac{1-(x+y)}{2(1-y)}$ if $3=60^x$ and $5=60^y$.

Aww, this problem looks easy peasy if we could solve it using the logarithm method as we can convert the exponential forms to logarithmic forms and get:

$\log_{60}3=x$ and $\log_{60}5=y$

Adding them up gives

$x+y=\log_{60}15$

Algebraically manipulating the above so we have

$-(x+y)=-\log_{60}15$

\begin{align*}1-(x+y)&=1-\log_{60}15\\&=\log_{60}60-\log_{60}15\\&=\log_{60}\dfrac{60}{15}\\&=\log_{60}4-(*)\end{align*}

And

$y=\log_{60}5$

\begin{align*}1-y&=\log_{60}60-\log_{60}5\\&=\log_{60}\dfrac{60}{5}\\&=\log_{60}12\end{align*}

Hence

$2(1-y)=2\log_{60}12=\log_{60}(12)^2=\log_{60}144-(**)$

Divide (*) by (**) we have:

$\dfrac{1-(x+y)}{2(1-y)}=\dfrac{\log_{60}4}{\log_{60}144}=\dfrac{\dfrac{\log_{12}4}{\log_{12}60}}{\dfrac{\log_{12}144}{\log_{12}60}}=\dfrac{\log_{12}4}{\log_{12}12^2}=\dfrac{\log_{12}4}{2}=\log_{12}2$

Convert this logarithm form into the exponential form again we have what we are desired:

$\large12^\frac{1-(x+y)}{2(1-y)}=2$

But, the question is, we are not allowed to solve using the help from logarithm method.

Now, take out a pencil, or a pen, to start scribble, this problem provides you the golden opportunity that you simply cannot pass up as this problem will promote you to think logically, abstractly, critically and creatively.

As usual, I will show you how I tackle the problem in my next blog post, see you!