AirForceOne said:
Is my answer correct? (the last part of my work)
Hmmmm......I'm sorry, but I don't follow what you attempted to do there.
Here's what I would do.
log<SUB>b</SUB> (xy<SUP>2</SUP>) + 2 log<SUB>b</SUB> (x/y) - 3 log<SUB>b</SUB> (y x<SUP>2/3</SUP>)
Each "multiplier" of a log is actually an exponent. I'll make that change first:
log<SUB>b</SUB> (xy<SUP>2</SUP>) + log<SUB>b</SUB> (x / y)<SUP>2</SUP> - log<SUB>b</SUB> (y x<SUP>2/3</SUP>)<SUP>3</SUP>
Next, I'd use the "power of a power rule." When you raise a power to a power, you multiply the exponents.
log<SUB>b</SUB> (xy<SUP>2</SUP>) + log<SUB>b</SUB> (x / y)<SUP>2</SUP> - log<SUB>b</SUB> (y<SUP>3</SUP> x<SUP>(2/3)*3</SUP>)
log<SUB>b</SUB> (xy<SUP>2</SUP>) + log<SUB>b</SUB> (x / y)<SUP>2</SUP> - log<SUB>b</SUB> (y<SUP>3</SUP> x<SUP>2</SUP>)
Now, I'll use some of the rules of logs to convert this to a single log expression:
log<SUB>b</SUB> [xy<SUP>2</SUP> * (x/y)<SUP>2</SUP>] - log<SUB>b</SUB> (y<SUP>3</SUP> x<SUP>2</SUP>)
log<SUB>b</SUB> [xy<SUP>2</SUP> * (x<SUP>2</SUP>/y<SUP>2</SUP>)] / (y<SUP>3</SUP>x<SUP>2</SUP>)
log<SUB>b</SUB> x<SUP>3</SUP> / (y<SUP>3</SUP> x<SUP>2</SUP>)
log<SUB>b</SUB> (x / y<SUP>3</SUP>)
There it is as a single log.
Or, I guess you could write it this way
log<SUB>b</SUB> x - 3 log<SUB>b</SUB> y
(Skeeter and Arthur....you are too fast for me!
)