SOLUTION: Please help! I've been show the rules, but I don't know how to apply them, could you please show me how? 3. Simplify the following expressions: a. X4 * X5 (4 & 5 are exp

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Please help! I've been show the rules, but I don't know how to apply them, could you please show me how? 3. Simplify the following expressions: a. X4 * X5 (4 & 5 are exp      Log On


   



Question 174244: Please help! I've been show the rules, but I don't know how to apply them, could you please show me how?
3. Simplify the following expressions:
a. X4 * X5 (4 & 5 are exponents, I don't know how to type that)
b. Z9/Z10 (9 & 10 are exponents, I don't know how to type that)
c. log3Y + log3Z (base 3)
d. log5Y - log5Z (base 5)

Found 2 solutions by nerdybill, josmiceli:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
3. Simplify the following expressions:
a. X4 * X5 (4 & 5 are exponents, I don't know how to type that)
x%5E4%2Ax%5E5
= x%5E%284%2B5%29
= x%5E9
.
b. Z9/Z10 (9 & 10 are exponents, I don't know how to type that)
Z%5E9%2FZ%5E10
= Z%5E%289-10%29
= Z%5E%28-1%29
OR
= 1%2FZ
.
c. log3Y + log3Z (base 3)
log3%28Y%29+%2B+log3%28Z%29
= log3%28Y%2AZ%29
= log3%28YZ%29
.
d. log5Y - log5Z (base 5)
log5%28Y%29+-+log5%28Z%29
= log5%28Y%2FZ%29
.
To review log rules see:
http://www.purplemath.com/modules/logrules.htm

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Think of the problem as if there were no exponents involved
(a)
x%5E4+=+x%2Ax%2Ax%2Ax
x%5E5+=+x%2Ax%2Ax%2Ax%2Ax so,
x%5E4%2Ax%5E5+=+x%2Ax%2Ax%2Ax%2Ax%2Ax%2Ax%2Ax%2Ax
This is x multiplied by itself 9 times, so
x%5E4%2Ax%5E5+=+x%5E9
Aha! You just add the exponents
You can do any problem this way & discover the rule
(b)
z%5E9+=+z%2Az%2Az%2Az%2Az%2Az%2Az%2Az%2Az
z%5E10+=+z%2Az%2Az%2Az%2Az%2Az%2Az%2Az%2Az%2Az

There's 1 more z in the denominator, I'm left with
z%5E9%2Fz%5E10+=+1%2Fz and I know
1%2Fz+=+z%5E-1
Aha!, I just do 9+-+10+=+-1
---------------------
As for logs, just do a very simple experiment
How about log%282%2C16%29+%2B+log%282%2C64%29
Big fact: ***** Logs are exponents!!! *****
So I'm adding 2 exponents here
The 1st term says: The exponent to the base 2 that gives me 16
The 2nd term says: The exponent to the base 2 that gives me 64
In the 1st case it's 4 because 2%5E4+=+16
In the 2nd case it's 6 because 2%5E6+=+64
So, now I,ve got
log%282%2C16%29+%2B+log%282%2C64%29+=+10
Remember that 10 is the sum of 2 logs, so it's a log.
Express it as log(2,?) or "log to the base 2 that gives me what?
If I raise 2%5E10+=+1024, so
log%282%2C16%29+%2B+log%282%2C64%29+=+log%282%2C1024%29
A better way to write this is:
log%282%2C2%5E4%29+%2B+log%282%2C2%5E6%29+=+log%282%2C2%5E10%29
Aha! Add the exponents
These can be brain-twisters, and it helps if you've
wrestled aligators, but there's no easy way out.