yay
remember that you can do anything to an equation as long as you do it to both sides
also
log(a)-log(b)=log(a/b)
and
if no base is stated, then probably base 10
also, [tex]log_a(b)=c[/tex] translates to [tex]a^c=b[/tex]
so
log(t-3)=log(17-4t)
minus log(17-4t) from both sides
log(t-3)-log(17-4t)=0
[tex]log(\frac{t-3}{17-4t})=0[/tex]
translate
remember, base 10
[tex]10^0=\frac{t-3}{17-4t}[/tex]
[tex]1=\frac{t-3}{17-4t}[/tex]
times both sides by 17-4t
17-4t=t-3
add 4t both sides
17=5t-3
add 3 both sides
20=5t
divide both sides by 5
4=t
t=4
