Answer :

gmany
[tex]\ln x+2\ln y-\ln z=\ln x+\ln y^2-\ln z=\ln\dfrac{xy^2}{z}\\\\Used:\ln a+\ln b=\ln(a\cdot b}\\\\\ln a-\ln b=\ln\dfrac{a}{b}\\\\n\ln a=\ln a^n[/tex]
TaeKwonDoIsDead

Answer:

[tex]ln(\frac{xy^{2}}{z})[/tex]


Step-by-step explanation:


3 rules of logarithms that we are going to use on this is:

  1. [tex]ln(a+b)=ln(a*b)[/tex]
  2. [tex]ln(a-b)=ln(\frac{a}{b})[/tex]
  3. [tex]bln(a)=ln(a)^{b}[/tex]

We can use the third rule to simplify the expression as:

[tex]ln(x)+2ln(y)-ln(z)\\=ln(x)+ln(y^{2})-ln(z)[/tex]

Simplifying the first 2 terms using first rule gives us:

[tex]ln(x*y^{2})-ln(z)[/tex]

Now using the second rule to further simplify and write as single logarithm gives us:

[tex]ln(\frac{xy^{2}}{z})[/tex]

Other Questions