Help ASAP show work please thanksss!!!!

Question

asked Mar 8, 2021 59.9k views

1 Answer

Tomislav Novoselec · Answer 1 · 2021-03-12T09:49:24+0000

Answer:

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

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:

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