Respuesta :
Answer:
true.
Step-by-step explanation:
[tex]b^t=\frac{n}{a}; \ = > \ t=log_b\frac{n}{a}; \ = > \ t=\frac{log\frac{n}{a}}{logb}.[/tex]
Answer:
true.
Step-by-step explanation:
[tex]b^t=\frac{n}{a}; \ = > \ t=log_b\frac{n}{a}; \ = > \ t=\frac{log\frac{n}{a}}{logb}.[/tex]