How do you evaluate log_169 (-13)log169(13)?

3 Answers
Aug 29, 2016

A log expression in this form is asking the question...

"what power of 169 will give -13?"

OR" What index of 169 will make -13?"

I

Aug 29, 2016

log_169 (-13)=1/2 + pi/(2ln(13))ilog169(13)=12+π2ln(13)i

Explanation:

For any Real value of xx, 169^x > 0169x>0, so cannot be equal to -1313

So to find a value for log_169(-13)log169(13) we need to consider Complex logarithms.

Note that e^(pi i) = -1eπi=1, so ln(-1) = pi iln(1)=πi

In general, if x < 0x<0 then ln(x) = ln abs(x) + pi iln(x)=ln|x|+πi

Use the change of base formula to find:

log_169 (-13)log169(13)

=ln(-13)/ln(169)=ln(13)ln(169)

=ln(-13)/ln(13^2)=ln(13)ln(132)

=(ln(13)+pi i)/(2 ln(13))=ln(13)+πi2ln(13)

=1/2 + pi/(2ln(13))i=12+π2ln(13)i

This is the principal value of the Complex logarithm.

Other values that satisfy 169^z = -13169z=13 are found by adding multiples of pi/(ln(13))iπln(13)i

Aug 30, 2016

1/2, (1+i(2n+1)pi)/(2 log 13), n = 0, +-1, +-2, +-3, ...

Explanation:

I think that I could make a compromising answer.

Use that, for a > 0, a^2=(+-a)^2 and

log_b a = log a/log b

Now,

log_169 (-13).

log(-13)/log 169

=log(-13)/log((+-13)^2)

=(log(-13)/log(+-13))/2

=1/2, for the negative sign, and, for the positive sign,

=log(13e^(i(2n+1)pi))/(2log 13), n = 0, +-1, +-2, +-3, ..,

using -1=cis ( odd integer multiple of pi)*

=( log 13+ i (2 n + 1) pi) / (2log 13), n = 0, +-1, +-2, +-3, ...

=1/2+i(2n+1)pi/(2log 13), n = 0, +-1, +-2, +-3,...

If students are not to be burdened, these questions could be

reserved for Extraordinary Talent Examinations, after doing good

home work on the answer....