logx^1/64 =-3/2

Question

logx^1/64 =-3/2

anonymous · Answer

Assuming I interpret your notation correctly,$$\log{x^{\frac{1}{64}}}=-\frac{3}{2}\rightarrow \frac{1}{64}\log{x}=-\frac{3}{2}$$So$$\log{x}=64 \times -\frac{3}{2}=-96$$Hence,$$x=e^{-96}$$

anonymous · Answer

This is log base 10 not log base e or natural log so the answer should be:

$$x=10^{-96}$$

anonymous · Answer

Doesn't have to be.

anonymous · Answer

Yeah it does unless you change the base for x.

log and ln are not equivalent.

put it in a TI-89 or some math program

anonymous · Answer

Well, it's a matter of definition.  Log is technically the notation for natural logarithm in general, with ln reserved for use as the natural logarithm of the magnitude of a complex number.  It's semantics.  Mathematicians use log.

anonymous · Answer

in maths log means base 10 not e unless it is mentioned.

anonymous · Answer

Whatever.  I have a degree + postgrad. in the subject.  I'm not arguing about it anymore.  Mathematicians use log for ln unless the base is made explicit.

anonymous · Answer

srry then.