38

Your work on this assignment is good.

Let me know if you have specific questions on any of these ideas.

......!!!!!!!!...................................

21:50:02 5.5.18. Exact value of log{base 3}{(8) * log{base 8}(9).

......!!!!!!!!...................................

RESPONSE --> The value of log base 3(8) is 2, but I cannot figure out what the value of the other log is. I know since this is multiplying two logs, then you add the two together.

log{base 3}(8) is not 2, since 3^2 = 9, not 8.

The key here is that log{base b}(x) = log(x) / log(b). The solution relies entirely on this rule.

.................................................

......!!!!!!!!...................................

21:50:44 log{base 3}(8) * log{base 8}(9) = log 8 / log 3 * log 9 / log 8 = log 9 / log 3 = log{base 3}(9). log{base 3}(9) is the power to which 3 must be raised to get 9, and is therefore equal to 2. Thus log{base 3}(8) * log{base 8}(9) = 2.

......!!!!!!!!...................................

RESPONSE --> I missed something somewhere here. So I will work on this area some more.

.................................................

......!!!!!!!!...................................

21:51:25 5.5.24. ln(2) = a, ln(3) = b. What is ln(2/3) in terms of a and b?

......!!!!!!!!...................................

RESPONSE --> ln 2/3 is = to ln 2- ln 3= a-b

.................................................

......!!!!!!!!...................................

21:51:29 ln(2/3) = ln(2) - ln(3) = a - b.

......!!!!!!!!...................................

RESPONSE --> ok

.................................................

......!!!!!!!!...................................

21:53:19 5.5.26. ln(0.5) in terms of a and b.

......!!!!!!!!...................................

RESPONSE --> ln .5= log base e .5 or ln 1/2= ln 1- ln 2= I am not sure where to go after this

.................................................

......!!!!!!!!...................................

21:53:29 Since ln(2) = a, and since ln(1) = 0, we have ln(.5) = ln(1/2) = ln(1) - ln(2) = 0 - a = -a.

......!!!!!!!!...................................

RESPONSE --> ok! I see now!

.................................................

......!!!!!!!!...................................

21:54:13 5.5.52. log{base 3}(u^2) – log{base 3}(v) as a single log.

......!!!!!!!!...................................

RESPONSE --> When I have subtraction, I can use the properties to divide and make this one log: log base 3 u^2/v

.................................................

......!!!!!!!!...................................

21:54:20 log{base 3}(u^2) – log{base 3}(v) = log{base 3}(u^2 / v).

......!!!!!!!!...................................

RESPONSE --> ok

.................................................

......!!!!!!!!...................................

21:54:32 5.5.68. Using a calculator express log{base1 / 2}(15)

......!!!!!!!!...................................

RESPONSE --> -3.9067

.................................................

......!!!!!!!!...................................

21:54:38 We get log{base 1/2}(15) = log(15) / log(1/2) = 1.176091259 / )-0.3010299956) = -3.906890595.

......!!!!!!!!...................................

RESPONSE --> ok

.................................................

......!!!!!!!!...................................

21:56:23 5.5.80. Express y as a function of x if ln y = ln(x + C).

......!!!!!!!!...................................

RESPONSE --> ln= log base e I am not sure where to go, I am really not good with ln

.................................................

......!!!!!!!!...................................

21:57:49 a = ln(b) means e^a = b, so y = ln(x+c) is translated to exponential form as (x+c) = e^y.

......!!!!!!!!...................................

RESPONSE --> ok, now that I have reviewed that a little more- exponential form, a= ln(b)= e^a=b y=ln(x+c)= (x+c)=e^y Very confusing. I have a long way to go with this and not much time to get there.

remember:

log{base b}(x) = y means the same thing as b^y = x.

Also, ln(x) is log{base e}(x), so ln(x) = y means that e^y = x.

It follows from this that a = ln(b) means e^a = b.

Finally

log{base b}(x) and b^x are inverse functions, so that log{base b}(b^x) = x and b^(log{base b}(x) ) = x, or in the case of the natural log function

ln(x) and e^x are inverse functions, so that e^(ln(x) ) = x and also ln(e^x) = x.

It also follows form this that a = ln(b) means e^x = e^(ln(b)) so that e^a = b.

.................................................

"