Show (log⁡n)(log⁡n)=2(log⁡n)(log⁡(log⁡n))I am having difficulty understanding how this follows.(log⁡n)(log⁡n)=2(log⁡n)(log⁡(log⁡n))=nlog⁡log⁡nWhich logarithmic identities are used to go through each equality?e.g. how do you first go from(log⁡n)(log⁡n)=2(log⁡n)(log⁡(log⁡n))and then to2(log⁡n)(log⁡(log⁡n))=nlog⁡log⁡n(The log base must be 2 or else this equality won't hold)

Question

Show (log⁡n)(log⁡n)=2(log⁡n)(log⁡(log⁡n))I am having difficulty understanding how this follows.(log⁡n)(log⁡n)=2(log⁡n)(log⁡(log⁡n))=nlog⁡log⁡nWhich logarithmic identities are used to go through each equality?e.g. how do you first go from(log⁡n)(log⁡n)=2(log⁡n)(log⁡(log⁡n))and then to2(log⁡n)(log⁡(log⁡n))=nlog⁡log⁡n(The log base must be 2 or else this equality won&#039;t hold)

enhebrevz · Accepted Answer

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Barbara Meeker · Accepted Answer

For any  1≠a∈R+,xy=aylogax,whenever LHS is defined.   Thus,  (log⁡n)log⁡n=2log⁡nlog⁡2(log⁡n)  So your equality follows if the logarithm here is taken in base 2 and not 10 , as you wrote...

RizerMix · Accepted Answer

Let y=log2log2⁡n⁡n Taking the logarithm (in base 2) of both sides log2⁡y=log2⁡nlog2⁡log2⁡n Now, remember that 2log2⁡n=n. Thus y=2log2⁡y=2log2⁡nlog2⁡log2 Also recall that abc=(ab)c. Thus y=2log2⁡nlog2⁡log2⁡n=(2log2⁡n)log2⁡log2⁡n=nlog2⁡log2⁡n