Question

Help ASAP show work please thanksss!!!!

Answer 1

`\displaystyle log_(1)/(2)(64)=-6`

Explanation:

Properties of Logarithms

We'll recall below the basic properties of logarithms:

`log_b(1) = 0`

Logarithm of the base:

`log_b(b) = 1`

Product rule:

`log_b(xy) = log_b(x) + log_b(y)`

Division rule:

`\displaystyle log_b((x)/(y)) = log_b(x) - log_b(y)`

Power rule:

`log_b(x^n) = n\cdot log_b(x)`

Change of base:

`\displaystyle log_b(x) = ( log_a(x))/(log_a(b))`

Simplifying logarithms often requires the application of one or more of the above properties.

Simplify

`\displaystyle log_(1)/(2)(64)`

Factoring
`64=2^6`.

`\displaystyle log_(1)/(2)(64)=\displaystyle log_(1)/(2)(2^6)`

Applying the power rule:

`\displaystyle log_(1)/(2)(64)=6\cdot log_(1)/(2)(2)`

Since

`\displaystyle 2=(1/2)^(-1)`

`\displaystyle log_(1)/(2)(64)=6\cdot log_(1)/(2)((1/2)^(-1))`

Applying the power rule:

`\displaystyle log_(1)/(2)(64)=-6\cdot log_(1)/(2)((1)/(2))`

Applying the logarithm of the base:

`\mathbf{\displaystyle log_(1)/(2)(64)=-6}`