Let a=log(base 12)2 Show that...?

Question

Let a=log(base 12)2 Show that log(base 6)16=4a/(1-a)

Anyone???

Answer 1

a = ln(2)/ln(12)
[i just picked ln() for convenience ... works for log to any base]

need to show:
4a = (1-a)*log_6(16)

4*ln(2)/ln(12) =?= [1-ln(2)/ln(12)] * ln(16)/ln(6)
4*ln(2) =?= [ln(12) - ln(2)]*ln(2^4)/ln(6)
4*ln(2) =?= [ln(12) - ln(2)] * 4ln(2) / ln(6)
ln(6) =?= ln(12) - ln(2)
which is true since ln(a/b) = ln(a) - ln(b)

Answer 2

i really dont see why you may have trouble, just use this rule, x = log base(y) z, then y^x = z, with the example you gave,

a = log(base 12)2, = 12^a = 2, so log(base 6)16=4a/(1-a) =

6^ (4a / 1-a) = 16, you already know what a is, so just evaluate a and plug into equation.

also to evaluate a in your calculator, just plug in log(2) / log (12) = a

Answer 3

a = log_12 2

Now
log_6 16 = log_6 (2^4) = 4 log_6 2

And
log_6 2 = log_12 2 / log_12 6

So
log_6 16 = 4 (log_12 2)/(log_12 6)

But
log_12 6 = log_12 (12/2) = log_12 12 - log_12(2) = 1 - log_12 (2)

so
log_6 16 = 4 (log_12 2)/(log_12 6)
= 4 (log_12 2)/(1 - log_12 2)
= 4a / (1-a)
QED

Answer 4

a=log(base 12)2
=> 4a=4log(base6)2*log(base12)6
=> 4a=log(base6)16*[log(base12)12 - log(base12)2]
=> 4a=log(base6)16*(1 - a)
=> log(base6)16 = 4a / (1-a)

Answer 5

4a/(1-a)=log6(16)
4a=log6(16)-alog6(16)
4a+alog6(16)=log6(16)
a(4+log6(16)=log6(16)
a=log6(16)/[4+log6(16)]
log6(16)[(4+log6(16)]=log12(2)
log6(16)=4log12(2)+log6(16)log12(2)
log12(16)/log12(6)=log12(16)+log12(16)log12(2)/log12(6)
[log12(16))[1-log12(2)]/log12(6)=log12(16)
log12(16)[log12(12)-log12(2)]=log12(16)log12(6)
log12(16)log12(6)=log12(16)log12(6)
0=0