## Thursday, April 23, 2015

### Evaluate without calculator or logarithm help

On my previous blog post(Evaluate $12^{\frac{1-(x+y)}{2(1-y)}}$), we're asked, without using the calculator and the help from logarithm, evaluate $\large12^\frac{1-(x+y)}{2(1-y)}$ provided $3=60^x$ and $5=60^y$.

Let do this problem as we're told, where we could not borrow help from calculator nor logarithms.

If we multiply the two given exponential equations, we get:

$3\cdot 5=60^x\cdot 60^y$

$3\cdot 5=60^{x+y}$

$3\cdot 5=5^{x+y}\cdot 12^{x+y}$

$3=5^{x+y-1}\cdot 12^{x+y}(*)$

whereas if we divide the two given exponential equations we obtain:

$\dfrac{3}{5}=\dfrac{60^x}{60^y}$

$\dfrac{3}{5}=60^{x-y}$

$\dfrac{3}{5}=5^{x-y}\cdot 12^{x-y}$

$3=5^{x-y+1}\cdot 12^{x-y}(**)$

By equating both equations (*) and (**), we see that

$5^{x+y-1}\cdot 12^{x+y}=5^{x-y+1}\cdot 12^{x-y}$

Upon simplifying gives

$\large12^{\frac{1}{2(1-y)}}=(5^{\frac{1}{b}})^{\frac{1}{2}}$ but note that $5^{\frac{1}{b}}=60$ so

$\large12^{\frac{1}{2(1-y)}}=(60)^{\frac{1}{2}}$

We're not finished yet, we need to raise both sides of the equation above by the quantity $1-(x+y)$, here we go:

$\large(12^{\frac{1}{2(1-y)}})^{1-(x+y)}=((60)^{\frac{1}{2}})^{1-(x+y)}$

$\large12^{\frac{1-(x+y)}{2(1-y)}}=\dfrac{(60)^{\frac{1}{2}}}{(60^{x+y})^{\frac{1}{2}}}$ note that $3\cdot 5=60^{x+y}$

$\therefore \large12^{\frac{1-(x+y)}{2(1-y)}}=\left(\dfrac{60}{3\cdot 5}\right)^{\frac{1}{2}}=4^{\frac{1}{2}}=2$