help logarithmic question?

Question

log 1/3 p < 0

logb 121 = 2


thanks

Answer 1

I'm not sure about the first one; I can't actually remember ever doing a logarithmic inequality.


You need to rewrite the second equation in exponential form before you can solve it.

Since b is the base and 2 is the exponent it should be rewritten this way:

b^2 = 121.

Take the square root of both sides.
b = 11 (or if you are considering more than just the principle root, b = -11 too).

Answer 2

the second one says b² = 121 (since the log is the exponent), so b = 11.

for the first, rewriting in exponential form, (1/3)^x = p, asking what values p has when x < 0. Well, if x = 0, p = 1. If x = -1, p = 3. If x = -2, p = 9. Just to make sure, if x = -1/2, p = √3. So for x < 0, p > 1.

Answer 3

Part one
log 1/3 p < 0
1/3 p < 10^0
1/3 p < 1
Multiplying both sides 3
p < 3

Part two
logb 121 = 2
b^2 = 121 = 11^2
b = +- 11

Answer 4

log 1/3 p < 0
1/3 p < 10^0
1/3 p < 1
p < 3

logb 121 = 2
b^2 = 121
b = sqrt(121)
b = +/- 11

Answer 5

Is logarithmic part of a branch of mathematics or is it its own branch? What is it exactly? I'm in grade/year 11 and I'm doing introductory calculus and also geometry and trigonometry. Will I be doing this? Sorry, I can't help you with you question. I'm interested in what it is though.