April 2020 1 93 Report
A equação 2log_2X - log_29 = 7^3log_7^2 tem como solução

a) x = -48
b) x= -27
c) x= 27
d) x = 36
e) x = 48
como estou resolvendo
log_2^.x^2 - log_2^.3^2 = 3.2
log_2 \frac{x^2}{3^2} = 6  log_2 \frac{x^2}{9} = 6  [tex]2^6 = \frac{x^2}{9}
64 =
9
x²= 576
x = 24

Please enter comments
Please enter your name.
Please enter the correct email address.
You must agree before submitting.

Lista de comentários


Helpful Social

Copyright © 2024 ELIBRARY.TIPS - All rights reserved.