MTAP Challenge

$90^a=2$ so $90^{1-a}=\dfrac{90}{90^a}=\dfrac{90}{2}=45$

Hence:

$A=45^{\frac{1-a-b}{2-2a}}=90^{(1-a)\frac{1-a-b}{2-2a}}=90^{\frac{1-a-b}{2}}=(90^{1-a-b})^{\frac{1}{2}}$

Also:

$90^{1-a-b}=\dfrac{90}{90^{a+b}}=\dfrac{90}{90^a\times 90^b}=\dfrac{90}{2 \times5}=9$

Then:

$A=9^{\frac{1}{2}}=3$

Taking logs base $45$ , and letting $\log$ represent $\log_{45}$ , we have that

$a = \dfrac{\log(2)}{\log(90)} = \dfrac{\log(2)}{1 + \log(2)} \Longrightarrow 1 - a = \dfrac{1}{1 + \log(2)}$ .

Also, $b = \dfrac{\log(5)}{1 + \log(2)}.$ . Thus

$\dfrac{1 - a - b}{2 - 2a} = \dfrac{1 - \log(5)}{2} = \dfrac{\log(9)}{2}$ .

we then have that

$45^{\frac{1 - a - b}{2 - 2a}} = (45^{\log(9)})^{\frac{1}{2}} = 9^{\frac{1}{2}} = \boxed{3}$ .

I converted $90^{a}=2$ to $\log_{90} 2 = a$ and and $90^{b} = 5$ to $\log_{90} 5 = b$ . This becomes a simple substitution problem. Eventually, you get $\displaystyle 45^{\frac{1}{2} \log_{45} 9}$ which is the same as $\displaystyle 45^{ \log_{45} 9^{\frac{1}{2}}}$ and applying the anti log law, you get 3.

90 ^(a + b) = 10

a + b = Log 10/ Log 90 = 0.51170721912445836072605926743669

a = Log 2 / Log 90 = 0.15403922195426357566083026916701

45^ 0.28860249407989831592687553274807 = 3