Find the minimum value of #log_2^2 2 + log_ 2^3 2^2 + log_ 2^4 2^3.............log_2^n 2^(n-1)#?

Find the minimum value of

#log_(2^2) 2 + log_ (2^3) 2^2 + log_ (2^4) 2^3.............log_(2^n) 2^(n-1)#

2 Answers
Apr 24, 2017

Answer:

The Reqd. Minimum Value =#(n-1)(1/n)^(1/(n-1)).#

Explanation:

We know the Change of Base Rule : #log_ca=log_ba/log_bc.#

Using #b!=1# as common base to all #log,# we have,

The Expression#=log_4 2+log_8 4+log_(16)8+...+log_(2^n) 2^(n-1),#

#=log_b2/log_b4+log_b4/log_b8+log_b8/log_b 16+...+log_b 2^(n-1)/log_b(2^n),#

#=log_b2/log_b2^2+log_b2^2/log_b2^3+log_b2^3/log_b 2^4+...+log_b 2^(n-1)/log_b(2^n),#

#=log_b2/(2log_b2)+(2log_b2)/(3log_b2)+(3log_b2)/(4log_b 2)+...+((n-1)log_b 2)/(nlog_b2),#

#=1/2+2/3+3/4+...+(n-1)/n....(ast)#

Now, we use the Arithmetic Mean-Geometric Mean Inequality, i.e.,

#AM ge GM#

Observe that the AM of #(n-1) +ve" Nos. of "(ast)# is,

#1/(n-1){1/2+2/3+3/4+...+(n-1)/n},# and, their GM is,

#{(1/2)(2/3)(3/4)...((n-1)/n)}^(1/(n-1))=(1/n)^(1/(n-1)).#

#:., 1/(n-1){1/2+2/3+3/4+...+(n-1)/n} ge (1/n)^(1/(n-1)), i.e.,#

# {1/2+2/3+3/4+...+(n-1)/n} ge (n-1)(1/n)^(1/(n-1)).#

Therefore, the Reqd. Minimum Value =#(n-1)(1/n)^(1/(n-1)).#

Enjoy Maths.!

Apr 24, 2017

Answer:

#sum_(k=1)^n log_(2^k) 2^(k-1) = n+sum_(k=1)^n(-1)^k1/k((n),(k))#

Explanation:

#log_(a^b) a^c = log a^c/(log a^b)=c/b#

so

#sum_(k=1)^n log_(2^k) 2^(k-1)= sum_(k=1)^n (k-1)/k = n-sum_(k=1)^n 1/k#

we know that

#1+x+x^2+ cdots + x^(n-1) = (x^n-1)/(x-1)#

so

#int_0^1 (x^n-1)/(x-1) dx = -int_0^1 ((1-y)^n-1)/y dy = - sum_(k=1)^n(-1)^k1/k((n),(k))#

so finally

#sum_(k=1)^n log_(2^k) 2^(k-1) = n+sum_(k=1)^n(-1)^k1/k((n),(k))#