Question

(Mackenzie-SP) Se (2^x. k^y+1. 5^t+3) (2^x−1. k^y. 5^t+1)^−1 = 150, então k vale:
a) 1
b) 2
c) 3
d) 4
e) 5

*OBS: O resultado tem que ser 3

Answer 1

$\displaystyle \sf \left(2^x\cdot k^{y+1}\cdot 5^{t+3}\right )\cdot \left(2^{x-1}\cdot k^{y}\cdot 5^{t+1} \right)^{-1} = 150 \\\\ \left(2^x \cdot k^{y}\cdot k \cdot 5^{t+1}\cdot 5^2\right)\cdot \left( 2^{x}\cdot 2^{-1} \cdot k^{y}\cdot 5^{t+1}\right)^{-1} =150 \\\\ 5^2\cdot k \cdot \left(2^x \cdot k^{y} \cdot 5^{t+1}\right)\cdot 2^{(-1)}^{\cdot (-1)} \cdot \left(2^x \cdot k^{y} \cdot 5^{t+1}\right)^{-1}=150$

$\displaystyle \sf 2\cdot 5^2\cdot k \cdot \frac{\left(2^x \cdot k^{y} \cdot 5^{t+1}\cdot \right)}{\left(2^x \cdot k^{y} \cdot 5^{t+1}\right)} = 150 \\\\\\ 2\cdot 25\cdot k = 150 \\\\ k = \frac{150}{50} \\\\ \huge\boxed{\ \sf k=3\ }\checkmark$

letra c