Question

Se+log+8(na+base+p)+=+-3/4+e+log+q(na+base+32)+=+3/5,+então+q+++1/p+é:
a)+21
b)+22
c)+23
d)+24
e)25

Answer 1

Resposta:

D) 24

Explicação passo-a-passo:

Se log 8(na base p) = -3/4 e log q(na base 32) = 3/5, então q + 1/p é:

a) 21

b) 22

c) 23

d) 24

e)25

$[/tex]</p><p>[tex]Log_p8=-\frac{3}{4}\\p=8^{-\frac{3}{4}}\\Log_{32}q=\frac{3}{5}\\q=32^{\frac{3}{5}}=(2^5)^{\frac{3}{5}}=8\\como\\q+\frac{1}{p}\\8+\frac{1}{8^{-\frac{4}{3}}}=8+(2^3)^\frac{4}{3}=8+2^{\frac{12}{3}}=8+2^4=8+16=24$