SOLUTION: So the question is as follows: "If log 2 + 0.301 and log 3 + 0.447, find each of the following using only these values and the properties of logarithms. Show your work. a. lo

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: So the question is as follows: "If log 2 + 0.301 and log 3 + 0.447, find each of the following using only these values and the properties of logarithms. Show your work. a. lo      Log On


   



Question 1012071: So the question is as follows:
"If log 2 + 0.301 and log 3 + 0.447, find each of the following using only these values and the properties of logarithms. Show your work.
a. log 6
b. log (2/3)
c. log 1.5
d. log 18"
my teacher stopped teaching the lessons because we have a limited time before exams, so she started assigning videos to watch. I watched the assigned ones, and many more but I just don't understand the question, I don't even really know what it's asking me to do.

Found 4 solutions by ankor@dixie-net.com, jim_thompson5910, ValorousDawn, MathTherapy:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
I think it should read:
"If log 2 = 0.301 and log 3 + 0.477, find each of the following using only these values and the properties of logarithms.
a. log 6
log(2*3) = log(2)+log(3), when multiplying two numbers you add their logs
therefore
log(6) = .301+.447 = .778
:
b. log (2/3) when dividing you subtract their logs
log(2/3) = log(2) - log(3)
log(2/3) = .301 - .477 = -.156
:
c. log 1.5
1.5 = (3/2)
log(3/2) = .477 - .301 = .176
:
d. log 18"
same as log%28%283%5E2%2A2%29%29, when raising a power you multiply log by the exponent
log(18) = 2(.477) + .301 = .954 + .301 = 1.255
:
Did this give you some idea of what is going here?

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assuming you meant to write this

log(2) = 0.301
log(3) = 0.477

I used a calculator to compute those above

--------------------------------------------------------------------

We will use these log rules

(1) log%28%28x%2Ay%29%29+=+log%28%28x%29%29%2Blog%28%28y%29%29

(2) log%28%28x%2Fy%29%29+=+log%28%28x%29%29-log%28%28y%29%29

(3) log%28%28x%5Ey%29%29+=+y%2Alog%28%28x%29%29

--------------------------------------------------------------------
a)

log(6) = log(2*3)

log(6) = log(2) + log(3) ... Use rule 1 (see above)

log(6) = 0.301 + 0.477

log(6) = 0.778

--------------------------------------------------------------------
b)

log(2/3) = log(2) - log(3) ... Use rule 2 (see above)

log(2/3) = 0.301 - 0.477

log(2/3) = -0.176

--------------------------------------------------------------------
c)

log(1.5) = log(3/2)

log(1.5) = log(3) - log(2) ... Use rule 2 (see above)

log(1.5) = 0.477 - 0.301

log(1.5) = 0.176

--------------------------------------------------------------------
d)

log(18) = log(2*9)

log(18) = log(2*3^2)

log(18) = log(2)+log(3^2) ... Use rule 1 (see above)

log(18) = log(2)+2*log(3) ... Use rule 3 (see above)

log(18) = 0.301 + 2*0.477

log(18) = 1.255


All answers are approximate

Answer by ValorousDawn(53) About Me  (Show Source):
You can put this solution on YOUR website!
Hi there. I'm sorry to hear about your situation, let me try my best.
Luckily, there's not a lot you have to know for these problems, but here's what you do have to know.
log%28ab%29=log%28a%29%2Blog%28b%29
log%28a%2Fb%29=log%28a%29-log%28b%29
log%28a%5Eb%29=b%2Alog%28a%29
With this, we can tackle the problems.
Six is really just 2 times 3, so log%286%29=log%282%2A3%29. We have from our first identity that log%28ab%29=log%28a%29%2Blog%28b%29, so log%282%2A3%29=log%282%29%2Blog%283%29. We know these values, and we can then use substitution. log%282%29%2Blog%283%29=0.301%2B0.447=0.748.

The second one is easier, as 2/3 is really straightforward. We use the identity log%28a%2Fb%29=log%28a%29-log%28b%29 here. We plug in a as 2 and b as 3 because they are the numerator and denominator respectively. We get log%282%2F3%29=log%282%29-log%283%29. Then we can use substitution of values to solve. log%282%29-log%283%29=0.301-0.447=-0.146.

Don't get tricked up on the third one because it's a decimal, remember that 1.5 is the same thing as 3/2 in fractional form. log%281.5%29=log%283%2F2%29. We use the same identity as above. log%283%2F2%29=log%283%29-log%282%29 log%283%29-log%282%29=0.447-0.301=0.146.

This one's a bit harder, but we just have to go back to the basics. Factorize, and you'll get 18=2%2A3%5E2. Therefore, log%2818%29=log%282%2A3%5E2%29. We can break it apart. We get log%282%2A3%5E2%29=log%282%29%2Blog%283%5E2%29, which due to log%28a%5Eb%29=b%2Alog%28a%29, becomes log%282%29%2Blog%283%5E2%29=log%282%29%2B2%2Alog%283%29. Substitute in your values, and you're done. log%282%29%2B2%2Alog%283%29=0.301%2B2%280.447%29=1.195

Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!
So the question is as follows:
"If log 2 + 0.301 and log 3 + 0.447, find each of the following using only these values and the properties of logarithms. Show your work.
a. log 6
b. log (2/3)
c. log 1.5
d. log 18"
my teacher stopped teaching the lessons because we have a limited time before exams, so she started assigning videos to watch. I watched the assigned ones, and many more but I just don't understand the question, I don't even really know what it's asking me to do.

a.
I guess you meant: log 2 = .0301 ; and log 3 = 0.477 i/o log 3 = cross%280.447%29
log 6
log 6 = log 2 * 3
log 6 = log 2 + log 3 ---- Applying log ab = log a + log b  
log 6 = .301 + .477 ------ Substituting given values of log 2 and log 3
highlight_green%28log+6+=+.778%29

b.
log+%282%2F3%29
log+%282%2F3%29+=+log+2+-+log+3 ---- Applying log+%28a%2Fb%29 = log a - log b
log+%282%2F3%29+=+.301+-+.477 ---- Substituting given values of log 2 and log 3
highlight_green%28log+%282%2F3%29+=+-+.176%29

c.
log 1.5 --------> log+%283%2F2%29
log+%283%2F2%29+=+log+3+-+log+2 ----- Applying log+%28a%2Fb%29 = log a - log b  
log+%283%2F2%29 = .477 - .301 --- Substituting given values of log 2 and log 3
highlight_green%28log+1.5+=+log+%283%2F2%29+=+.176%29

d.
log 18 -------> log+3%5E2+%2A+2+
log 18 = log+3%5E2+%2B+log+2 ----- Applying log ab = log a + log b
log 18 = 2 log 3 + log 2 -----Applying log+a%5Eb = b log a
log 18 = 2(.477) + .301 ------ Substituting given values of log 2 and log 3
log 18 = .954 + .301
log 18 = highlight_green%281.255%29