Let \(\log_{4}3=x\) . Then \(\log_{2}27=kx\). Find k .
The first says that 4^x = 3
The second says that 2^(kx) = 27
Re-writing the first, we have
[ 2^2]^x = 3
[2^x]^2 = 3 cube both sides
( [2^x]^2 )^3 = 27
2^(6x) = 27
Which implies that
2^(6x) = 2^(kx)
Which implies that k = 6
Oh! This makes sense!
I couldn't figure it out earlier!
Thanks!
