Log3(2)=a. Demonstrate that log16(24)=(1+3a)/4a.

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log16(24)=log3(24)/log3(16)=log3(2^3*3)/log3(2^4)=

=[log3(2^3)+log3(3)]/4log3(2)=[3log3(2)+1]/4log3(2)

From hypothesis, we know that log3(2)=a. We'll substitute in the last expression found after log16(24) calculus.

[3log3(2)+1]/4log3(2)=(3a+1)/4a q.e.d.

Log3(2)=a=>2=3^a...(1)

Let log16(24)=x=>24=16^x...(2)

Let us find x in terms of a:

From(2): 24=16^x=> 3*2^3 =>(2^4)^x, as 16=2^4 and 8=2^3,

3*(2)^3=(2^4)^x. From (1) substitute, 2=3^a

3*(3^a)=(3^a)^(4x)

3^(a+1)=(3)^4ax Equating the powers of the same base,3 we get:

a+1=4ax. Therefore, **x= (1+a)/4a =RHS**.

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