Simplifying Logs: -3 log_2(x) - 2 log_2(y) + (1/2) log_2(z)

[RB67]

Instructions: Use the properties of logarithms to write each expression as the logarithm of one quantity.

Problem:

I need some help getting started with this one. Thanks.

 

[pka]

\(\displaystyle \L - 3\log _2 \left( x \right) - 2\log _2 \left( y \right) + \frac{1}{2}\log _2 \left( z \right) = \log _2 \left( {x^{ - 3} } \right) + \log _2 \left( {y^{ - 2} } \right) + \log _2 \left( {\sqrt z } \right)\)

 

[RB67]

OK after that I got :

Is that correct?

 

[Deleted member 4993]

Looks good to me....

 

[RB67]

One more question for tonight.

Just help getting started on this one would be appreciated.

 

[tkhunny]

Why is it different from the first one? Use the rules.

 

[RB67]

I tried and this is what I got:

log(xy + y^2) / log(xz + yz)(z)

It doesn't look right to me though.

 

[Deleted member 4993]

I tried and this is what I got:

log(xy + y^2) / log(xz + yz)(z)

It doesn't look right to me though.

It is not.

Use
the following rules:

log A - log B = log(A/B)................the bases of logmust be same.

so

log(xy+y^2) - log(xz + yz) = log[(xy+y^2)/(xz+yz)} = log[{y(x+y)}/{z(x+y)}] = log(y/z)

Now continue....

Start a new thread with a new problem

 

[morson]

Assuming z, x and y > 0 and x \(\displaystyle \not=\ -y\):

\(\displaystyle \L\ log(xy + y^2) - log(xz + yz) + log(z)\)

= \(\displaystyle \L\ log(y(x + y)) - log(z(x + y)) + log(z)\)

= \(\displaystyle \L\ log(\frac{yz(x + y)}{z(x + y)}\)\)

Cancel.

= \(\displaystyle \L\ log(y)\)