Question

Se 1989^a =13 e 1989^b= 17, o valor de 117 ^ 1-a-b/2(1-b) é? Obs: Só pode ser feita por potenciação!

Answer 1

Como 1989ᵃ = 13, então:

$1989^a = 13 \\\\a\log(1989) = \log(13) \\\\a = \dfrac{\log(13)}{\log(1989)}$

Como 1989ᵇ = 17, então:

$1989^b = 17 \\\\b\log(1989) = \log(17) \\\\b = \dfrac{\log(17)}{\log(1989)}$

Substituindo no numerador e no denominador, temos:

$1 - a - b = 1 - \dfrac{\log(13)}{\log(1989)} - \dfrac{\log(17)}{\log(1989)} = \dfrac{\log(1989) - \log(13) - \log(17)}{\log(1989)} \\\\\\2(1 - b) = 2 \left(1- \dfrac{\log(17)}{\log(1989)} \right) = 2\left( \dfrac{\log(1989) - \log(17)}{\log(1989)}\right)$

Dividindo ambos os termos, temos:

$\dfrac{1-a-b}{2(1-b)} = \dfrac{\dfrac{\log(1989) - \log(13) - \log(17)}{\log(1989)}}{ \dfrac{2(\log(1989) - \log(17))}{\log(1989)}} = \\\\\\\dfrac{1-a-b}{2(1-b)} = \dfrac{\log(1989) - \log(13) - \log(17)}{2(\log(1989) - \log(17))} = \\\\\\\dfrac{1-a-b}{2(1-b)} = \dfrac{\log \left( \dfrac{1989}{13} \right) - \log(17)}{2\log \left( \dfrac{1989}{17} \right)} = \dfrac{\log(153) - \log(17)}{2\log(117)}$

Portanto, podemos concluir que:

$\\\\\\\dfrac{1-a-b}{2(1-b)} = \dfrac{\log \left( \dfrac{153}{17} \right)}{2\log(117)} = \dfrac{\log(9)}{2\log(117)} = \dfrac{2\log(3)}{2\log(117)}=\dfrac{\log(3)}{\log(117)}$

Elevando a 117, temos:

$117^{\left( \dfrac{1-a-b}{2(1-b)} \right)} = 117^{\left( \dfrac{\log(3)}{\log(117)} \right)} = 3$