Question

What is the value of log0.5 16

Answer 1

You mean ?

`log_0_._5 (16) =-4`

or

You mean?

`log(0.5)^(16)=-11.09035489`

Explanation:

If you trying to solve:

`log_0_._5(16)`

Then use the definition of the base of a log which is:

`log_xy=z\Rightarrow x^z=y`

`0.5^z=16`

Rewrite 0.5 as:

`0.5=(1)/(2)`

As you may know:

`2^4=16`

`then\\\\z=-4`

Let's use the negative exponent propierty in order to verify the result:

`a^(-n)=(1)/(a^n)`

`((1)/(2) )^(-4)=(1)/((1^4)/(2^4) ) =2^4=16`

If you trying to solve:

`log(0.5)^(16)`

Then use reduction of power propierty:

`log(x^y)=y log(x)`

Therefore:

`log(0.5)^(16) =16 *log(0.5)=16*(-0.6931471806)=-11.09035489`

Answer 2

the answer:
the full question is as follow
What is the value of log0.5 ^16

first of all, we should know some logarithm property
for example:
log a^p = p log a, for all a a positive number, an p a real number

therefore, log0.5 ^16 = 16log0.5, and log0.5 = -0.69,
finally
log0.5 ^16 = 16log0.5=16*(-0.69)= -11