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

Question

3834 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-29T15:39:45

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

Answer 2 · 2016-08-29T18:50:30

#log_169 (-13)=1/2 + pi/(2ln(13))i#

For any Real value of #x#, #169^x > 0#, so cannot be equal to #-13#

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

Note that #e^(pi i) = -1#, so #ln(-1) = pi i#

In general, if #x < 0# then #ln(x) = ln abs(x) + pi i#

Use the change of base formula to find:

#log_169 (-13)#

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

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

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

#=1/2 + pi/(2ln(13))i#

This is the principal value of the Complex logarithm.

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

Answer 3 · 2016-08-30T06:45:13

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

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....