Answer:
You mean ?
[tex]log_0_._5 (16) =-4[/tex]
or
You mean?
[tex]log(0.5)^{16}=-11.09035489[/tex]
Step-by-step explanation:
If you trying to solve:
[tex]log_0_._5(16)[/tex]
Then use the definition of the base of a log which is:
[tex]log_xy=z\Rightarrow x^z=y[/tex]
[tex]0.5^z=16[/tex]
Rewrite 0.5 as:
[tex]0.5=\frac{1}{2}[/tex]
As you may know:
[tex]2^4=16[/tex]
[tex]then\\\\z=-4[/tex]
Let's use the negative exponent propierty in order to verify the result:
[tex]a^{-n}=\frac{1}{a^n}[/tex]
[tex](\frac{1}{2} )^{-4}=\frac{1}{\frac{1^4}{2^4} } =2^4=16[/tex]
If you trying to solve:
[tex]log(0.5)^{16}[/tex]
Then use reduction of power propierty:
[tex]log(x^y)=y log(x)[/tex]
Therefore:
[tex]log(0.5)^{16} =16 *log(0.5)=16*(-0.6931471806)=-11.09035489[/tex]
