form inequality for approximations and contradictions
log(a)(b) = 1/log(b)(a)
log(a)(x) is the inverse of
a^x
a^log(a)(x) = x
log(a)(b) < c –> b < a^c
3(root3) < 4
27 < 16
contradiction
therefore 3(root3)>4
log_a_b < c –>
b < a^c
Comparing logs:
Change bases, form inequalities, make approximations where appropriate, form inequalities by considering the approximations
Log(a) < 0 when
a < 1
log(a)(b) =
log(c)(b) / log(c)(a) === 1/log(b)(a)
Log1 =
0
a^ (log(a)x) =
log(a)(a^x) = x
change bases on largest log values and
find contradiction
a^loga(b) =
b
log(2)(3) > 1.5 therefore
2^1.5 < 3
logab < c
Raise both sides to the base of the log, a
a^logab < a^c
LHS = b
Therefore a^c > b