ChristianFreyre
ChristianFreyre ChristianFreyre
  • 16-10-2016
  • Mathematics
contestada

i need to evaluate
log base 3 (1/162)

Respuesta :

apologiabiology
apologiabiology apologiabiology
  • 16-10-2016
log(a*b)=log(a)+log(b)
and
logₐbˣ=xlogₐb
also
x^-m=1/(x^m)
and
logₐaˣ=x

so
[tex]log_3 \frac{1}{162} [/tex]=y
[tex]log_3 \frac{1}{2*3^4} [/tex]=y
[tex]log_3 \frac{1}{2} + log_3 \frac{1}{3^4} [/tex]=y
[tex]log_3 \frac{1}{2} + log_3 (3^{-4}) [/tex]=y
[tex]log_3 \frac{1}{2} -4 [/tex]=y
the result is
[tex](log_3 \frac{1}{2})-4 [/tex]
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