Q: Prove- 2/log_8a - 4/log_2a= 1/log_4a

Question

i tried dividing 1/log_4a by 2/log_4(8) - 1/log_4a by 4/log_4(2) to get
log_4(8)/2.log_4a - log_4(2)/4.log_4a
=log_4(8)/log_4a^2 - log_4(2)/log_4a^4
=1.5/log4a^2 - .5/log4a^4
=1/....
I don't know how to deal with the denominator or if any of that right
Thanks!

Answer 1

log_2(a) = 2log_4(a) = 3log_8(a)
So, you have

2/(1/3 log_2(a)) - 4/log_2(a) = 1/(1/2 log_2(a))

1/log_2(a) (6 - 4) = 2/log_2(a)
6-4 = 2

Or, recall that

log_b(a) = 1/log_a(b)
So, you have

2log_a(8) - 4log_a(2) = log_a(4)
But, since 8=2^3 and 4=2^2, that means
6log_a(2) - 4log_a(2) = 2log_a(2)
done