THREE MORE LOG RULES:

1. The log of it's own base ALWAYS 1. For example Log base 5 of 5 is 1, Log base 20 of 20 is 1, and Log of 10 is 1 since Log with no base written is always base 10. So LN(e) equals 1, because the base of LN is always e.

2. Any Log of 1 is ALWAYS Zero. Log(base 6) of 1 = 0. Log(1) =0, LN1 = 0.

3. You CANNOT Log a negative number. This is invalid and results in "ERROR"

SOLVE:

Find Log base x of Z if: Log base b of X = 2 and Log base b of Z = 5

First, write down what is being asked of you, Find Log base x of Z means:

X to what power equals Z? so the question is: X^(?) = Z and its given that

b^2 = X and that b^5 = Z. So if we log both sides of the Question (X^(?) = Z) we get:

Log[X^(?)] = Log(Z), using the Log Power Rule we get (?)LogX = LogZ, solving we get (?) = Log(Z)/Log(X). Now we substitute and we get (?) = Log(b^5)/Log(b^2). Using the Log Power Rule again, (once on top and once on the bottom) we get

(?) = 5Logb / 2Logb. From here the Logb's cancle giving: 5/2

SOLVE:

5Log base 5 of 6 + Log base 5 of 7

Here we CANNOT combine these using the addition/product rule yet because there is a 5 in front of the first term, (5Log56). So we first use the power rule in reverse to get rid of the 5 in front.

Now we have Logbase5 of 6^5 + Log base 5 of 7. From here we can use the addition rule since there is no number in front of eiother term. The problem now terns into Logbase5(6^5*7).

So essentially this means 5 to what power equals (6^5*7)? or 5^x = 6^5*7. Well 6^5 = 7776 and 7776* 7 = 54432. So we have 5^x = 54432. Taking Log of both sides we have:

Log5^x = Log(54432) . Using the Power Rule xLog5 = Log 54432, so

X = Log(54432)/Log5 = 6.77548 which checks out since:: 5^(6.77548) = 54432

Express the following in terms of only A and B.

LN(4th root of 2/3) where A = Ln2 and B = Ln3. Remember that the square root of a number is the same as the number to the 1/2 power. So cube root would be to the 1/3 power and our problem says 4th root so it means 2/3 to the 1/4 power. From Ln(2/3)^1/4 we use the power rule and we now have (1/4)Ln(2/3). From here we use the Subtraction/Division Rule but make sure and keep the 1/4 outside of everything: (1/4)(Ln2 - Ln3). Now Substite and you have (1/4)(A - B).

So the question expressed only in terms of A and B is: (A-B)/4

SOLVE:

2LN(Square root of e)

Well square root means to the 1/2 power, so the question can be re-written as 2LNe^1/2. Using the Power Rule in reverse we put the 2 on the outside back into power, so we have

LN(e^1/2)^2, from here remember that when you have a power to a power you have to multiply them, so (1/2)*(2) = 1 so we really have LN(e^1) or just LNe and remember that the base of LN is e and the rule that Log or LN of it's own base is ALWAYS 1. But here's a way of proving the rule in this problem. We have LNe = x where the base is also e. So really we have e^x = e. Well from here (even though it's obvious the answer can only be 1) we can log both sides and get Log(e^x) = Log(e), then using the power rule and solving we would have xLoge = Loge and x = Loge/Loge which equals 1. This can also be checked by calculator: (2)*ln(sqrt(e) = 1

Answer = 1.

SOLVE:

Log (base 5 of the 3rd root of 25)

Third root means to the 1/3 power, so we have Logbase5(25)^1/3. So the equation we have is Logbase5(25)^1/3 = X, this gives 5^X = 25^1/3. From here we log both sides and get

Log5^X = Log(25^1/3). Notice now we have gotten rid of the base five. Using the Power Rule of both sides we have: XLog5 = 1/3Log25, so X = 1/3(Log25)/Log5. But from here you should notice that Log25 can be broken down into Log5 + Log5 so we really have:

1/3[(Log5 +Log5)/Log5). After canceling we just have: 1/3(1 + 1) which is 2/3.

A second shorter way to solve this but maybe harder to see is that once you have

5^X = 25^1/3 the 25 can be broken down into 5*5 which is also 5^2 so you really have 5^X equaling (5^2)^1/3 and remember when you have a power to a power you multiply so 25^1/3 is really 5^2/3. So you just have 5^X = 5^2/3 so the answer must be 2/3.