A problem on Logarithms

[Guest]

Hi,
Here is the problem that I want to simplify:

Here's my attempt to simplify it:

Or if you perfer the above in "TeX":

Is my answer correct? (the last part of my work)
Sorry if it looks like I have no clue what I am doing

Thanks!

 

[skeeter]

I get \(\displaystyle \L \log_b{\left(\frac{x}{y^3}\right)}\)

 

[arthur ohlsten]

log[xy^2]+2 log [x/y] -3 log [xy^2/3]
log[xy^2] + log [x/y]^2 -log[xy^2/3]^3
log [xy^2 +log x^2/y^2 -log [x^3y^2]
log[ [ xy^2 x^2] / [y^2x^3y^2] ]
log [ x^3y^2/x^3y^4 ]
log 1/y^2
-2logy answer
Arthur

 

[Mrspi]

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! )

 

[Guest]

Thanks! I totally forgot that the number in front can be used as an exponential. So Spi's and Skeeter's answer must be correct...
Thanks