Can you answer in 15 seconds? (Part 1)

Note that:

$27^{\dfrac{-x}{3}} + 81^{\dfrac{1 - x}{4}}$

$= 3^{\dfrac{-3x}{3}} + 3^{\dfrac{4(1 - x)}{4}}$

$= 3^{-x} + 3^{1 - x}$

$= 3^{-x}( 3^{-x + x} + 3^{1 - x + x})$

$= \dfrac{1}{3^x}(3^0 + 3^1)$

$= \dfrac{1}{3^x}(1 + 3)$

$= \dfrac{1}{3^x}(4)$

$= \dfrac{4}{3^x}$

So $a + b = 4 + 3 = 7 . \square$

Note that: ${27}^{\frac{-x}{3}}+{81}^{\frac{1-x}{4}}=\frac{1}{{27}^{\frac{x}{3}}}+\frac{1}{{81}^{\frac{x-1}{4}}}=\frac{1}{{3}^{\frac{(3)x}{3}}}+\frac{1}{{3}^{\frac{(4)(x-1)}{4}}}=\frac{1}{{3}^{x}}+\frac{1}{{3}^{x-1}} \rightarrow \frac{1}{3^x}+\frac{3}{3^x}=\frac{4}{3^x}$

$4+3=7$

* $27^{(-x/3)} + 81^{(1-x)/4} = 3^{-3x/3} +3^{\frac{4(1-x)}{4} } = 3^{-x} + 3^{1-x} =\frac{1}{3^{x}} + \frac{3}{3^{x}} =\frac{1+3}{3^{x}} = \frac{4}{3^{x}}$ *

Note that: $27^{\frac{-x}{3}} + 81^{\frac{1-x}{4}}$ $=3^{\frac{-3x}{3}} + 3^{\frac{4(1-x)}{4}}$ $=3^{-x} + 3^{1-x}$ $=\frac{1}{3^{x}} + \frac{3}{3^{x}}$ $=\frac{1+3}{3^{x}}$

Comparing with given form:

$\frac{1+3}{3^{x}} = \frac{a}{b^{x}}$ $a = 1+3; b = 3$

So, a+b=7

After factoring we get 4/3^x as shown in other solutions. But won't seven be the minimum value of a+b since b is dependent on the value of x?

easy ha question

I just guessed

$\large \begin{aligned} 27^{- \frac {x}{3} } + 81^{ \frac {1-x}{4} } &= \sqrt[3]{27} ^{-x} + \sqrt[4]{81}^{-x+1} \\ \\ &= \frac{1}{\sqrt[3]{3^3}^x} + \sqrt[4]{3^4}^{-x+1} \\ \\ &= \frac{1}{3^x} + 3^{-x+1} \\ \\ &= \frac{1}{3^x} +\frac{3}{3^x} \\ \\ &= \frac{4}{3^x} \\ \\ a&=4 \\ \\ b&=3 \\ \\ a+b&=4+3=\boxed{7} \end{aligned}$

3^3×(-x)÷3+3^4 (1-x)÷4 1/3^x+3/3^x 4/3^x 4+3 7

Wow this was easy