Question

How do I solve log(2)t^3-1=-3?

Answer 1

Assuming you mean
$log_2(t^3-1)=-3$
remember
$log_a(b)=c$ translates to $a^c=b$
translate

$log_2(t^3-1)=-3$ translates to $2^{-3}=t^3-1$
$\frac{1}{2^3} =t^3-1$
$\frac{1}{8} =t^3-1$
add 1 to both sides
$\frac{9}{8} =t^3$
cube root both sides
$\frac{ \sqrt[3]{9} }{2}=t$