shamrock2004
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Use the properties of logarithms to expand the expression as a sum, difference, and/or constant multiple of logarithms. (Assume the variable is positive.)
ln z(z − 1)^8, z > 1

Respuesta :

semsee45

Answer:

[tex]\ln z + 8\ln (z-1)[/tex]

Step-by-step explanation:

[tex]\ln z(z-1)^8[/tex]

Using the product rule: [tex]\ln(xy)=\ln x + \ln y[/tex]

[tex]\implies \ln z + \ln (z-1)^8[/tex]

Using the power rule:  [tex]\ln x^n=n \ln x[/tex]

[tex]\implies \ln z + 8\ln (z-1)[/tex]
  • ln(ab)=lna+lnb

[tex]\\ \rm\Rrightarrow lnz(z-1)^8[/tex]

[tex]\\ \rm\Rrightarrow lnz+ln(z-1)^8[/tex]

  • lna^b=blna

[tex]\\ \rm\Rrightarrow lnz+8ln(z-1)[/tex]

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