# Prove that log 16 24*4*log 3 2 = 1 + 3*log 3 2

hala718 | High School Teacher | (Level 1) Educator Emeritus

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(log 16 24)* 3log 3 2 = 1 + 3log 3 2

We know that:

log a b = log c b/ log c a

==> log 16 24 = log 3 24/log 3 16

= log 3 (3*2^3)/ log 3 (2^4)

=[ log 3 (3) + 3 log 3 (2)] / 4 log 3 (2)

=(1 + 3log 3 2)/4log 3 2

Now multiply by 4log 3 (2):\

==>   [l1+ 2log 3(2)/4log 3 (2)]*4log 3 2

==> (1+ 2 log 3 (2)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Supposing that 16 and 3 are the bases of logarithms, we'll note

log3(2)=a.

We'll transform the base 16 into the base 3:

log16(24)=log3(24)/log3(16)

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

We'll use the product rule of logarithms:

log16(24)=[log3(2^3)+log3(3)]/4log3(2), where log 3 (3) = 1

log16(24)=[3log3(2)+1]/4log3(2) (*)

Let's recall that we've noted log3(2)=a.

We'll substitute log3(2)=a in expression (*).

(3a+1)/4a=(3a+1)/4a

The expression from the left side is identical with the expression from the right side.

jeyaram | Student, Undergraduate | (Level 1) Valedictorian

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log 16 24*4*log 3 2 = 1 + 3*log 3 2

L.H.S=log 16 24*4*log 3 2

=log 3 24/log 3 16 * 4log 3 2

=(log 3 3+log 3 8)/log 3 16 * 4log 3 2

=(1+log 2 8/log 2 3)/(log 2 16/log 2 3) * 4log 3 2

=(log 2 3+3)/4 * 4log 3 2

=(1/log 3 2  + 3)log 3 2

=(1+3log 3 2)

R.H.S

neela | High School Teacher | (Level 3) Valedictorian

Posted on

LHS :

log16(24) * 4 log3 (2)  = {log 3 (24)/ log3 (16} (log 3 (2^4) as log a (b) = log k(a)/logk(b)

=log3 (24)/log3 (16) } (log 3(16)} = log3 (24).

RHS = 1+ 3log3 (2) = log3 (3)+ log3 (2^3) , as m loga = loga^m

= log3 (3) + log3 (8)

= log3 (3*8) , as loga+logb = logab

= log3 (24) which is LHS.