SOLUTION: Can you please try to show your steps on how you got the following answers? Given log[a](5)=2.3 and log[a](3)=1.6, fill in the table below with the appropriate values. x

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Can you please try to show your steps on how you got the following answers? Given log[a](5)=2.3 and log[a](3)=1.6, fill in the table below with the appropriate values. x       Log On


   



Question 594466: Can you please try to show your steps on how you got the following answers?
Given log[a](5)=2.3 and log[a](3)=1.6, fill in the table below with the appropriate values.
x 15 ; 9 ; 5/3 ; 5a ; (3)/(a^2)
log[a](x)

Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
log%28a%2C+%285%29%29=2.3
log%28a%2C+%283%29%29=1.6
Plus,
log%28a%2C+%28a%29%29+=+1 (This is always true, no matter what "a" is, so it does not need to be told this.)
you have been asked to find various other base a logarithms. The "trick" to these is to use algebra and/or properties of logarithms to rewrite the desired logs in terms of the logs you already know.

So for
log%28a%2C+%2815%29%29
we want to rewrite the 15 in terms of 5's, 3's and/or a's. I hope that you can see that 15 is 5*3. Replacing the 15 with 5*3 we get:
log%28a%2C+%285%2A3%29%29
Now we use a property of logarithms for logs of a product, log%28x%2C+%28p%2Aq%29%29+=+log%28x%2C+%28p%29%29+%2B+log%28x%2C+%28q%29%29, we can separate the 5 and 3:
log%28a%2C+%285%29%29+%2B+log%28a%2C+%283%29%29
Now that we have the log of 15 expressed in terms of logs of 5 and 3. We can now use the given values:
2.3 + 1.6
which simplifies to 3.9. So log%28a%2C+%2815%29%29+=+3.9

log%28a%2C+%289%29%29
For 9 we could use either 3*3 or 3%5E2. I'll use the later one so you can see another property in use:
log%28a%2C+%283%5E2%29%29
Using a property for logs of a power, log%28x%2C+%28p%5Eq%29%29+=+q%2Alog%28x%2C+%28q%29%29 we can separate the exponent from the 3:
2%2Alog%28a%2C+%283%29%29
Replacing the log with its given value we get:
2 * 1.6
which simplifies to
3.2
So log%28a%2C+%289%29%29+=+3.2

log%28a%2C+%285%2F3%29%29
5/3 is already expressed in terms of 5's and 3's. Using a property for logs of quotients, log%28x%2C+%28p%2Fq%29%29+=+log%28x%2C+%28p%29%29+-+log%28x%2C+%28q%29%29 we get:
log%28a%2C+%285%29%29+-+log%28a%2C+%283%29%29
Replacing the logs with their given values we get:
2.3 - 1.6
which simplifies to
0.7
So log%28a%2C+%285%2F3%29%29+=+0.7

log%28a%2C+%283%2Fa%5E2%29%29
First we'll use the property for quotients:
log%28a%2C+%283%29%29+-+log%28a%2C+%28a%5E2%29%29
and then the property for powers (on the second log):
log%28a%2C+%283%29%29+-+2%2Alog%28a%2C+%28a%29%29
Now we can replace the logs with their known values.
1.6 - 2*1
which simplifies
1.6 - 2
-0.4
So log%28a%2C+%283%2Fa%5E2%29%29+=+-0.4