How do you condense #6log_2 (2/3)+2log_2(1/6)-4log_2(2/9)#?

1 Answer
Mar 23, 2016

Answer:

The expression can be condensed to #0#.

Explanation:

We can use the following logarithmic laws:

#[1]" " log_a x + log_a y = log_a (x * y)#

#[2]" " log_a x - log_a y = log_a (x / y)#

#[3]" " r * log_a x = log_a (x^r)#

As all logarithmic expressions have the same basis, all those laws can be applied without any further preparation.

Thus, we have:

#6 log_2(2/3) + 2 log_2(1/6) - 4 log_2(2/9) #

# stackrel("[3] ")(=) log_2( (2/3)^6) + log_2( (1/6)^2 ) - log_2( (2/9)^4)#

# stackrel("[1] ")(=) log_2( (2/3)^6* (1/6)^2 ) - log_2( (2/9)^4)#

# stackrel("[2] ")(=) log_2( ((2/3)^6* (1/6)^2 ) / (2/9)^4)#

Now, the expression is condensed. However, it can still be simplified:

#((2/3)^6* (1/6)^2 ) / (2/9)^4 = (2^6 / 3^6 * 1 / 6^2) / (2^4 / 9^4)#

... dividing by a fraction is the same thing as multiplying by the reciprocal...

#" " = (2^6 * 9^4) / (3^6 * 6^2 * 2^4)#

#" " = (2^6 * (3^2)^4) / (3^6 * (2*3)^2 * 2^4)#

... use the power rule #(a^m)^n = a^(m*n)#:

#" " = (2^6 * 3^8) / (3^6 * 2^2 * 3^2 * 2^4)#

... use the power rule #a^m * a^n = a^(m+n)#:

#" " = (2^6 * 3^8) / (color(blue)(3^6) * color(orange)(2^2) * color(blue)(3^2) * color(orange)(2^4))#

#" " = (2^6 * 3^8) / (color(blue)( 3^8) * color(orange)(2^6))#

... the numerator and the denominator can be cancelled completely...

#" " = 1#

Thus, in total, we have:

# log_2( ((2/3)^6* (1/6)^2 ) / (2/9)^4) = log_2 (1) = 0#,

as for any basis #a#, it holds #log_a(1) = 0# .