If I understand the question properly, we have:
#log8^x=p#
And we wish to express #log2^x# in terms of #p#.
The first thing we should note is that #log8^x=xlog8#. This follows from the following property of logs:
#loga^b=bloga#
Essentially, we can "bring down" the exponent and multiply it by the logarithm. Similarly, using this property on #log2^x#, we get:
#log2^x=xlog2#
Our problem is now boiled down to expressing #xlog2# (the simplified form of #log2^x#) in terms of #p# (which is #xlog8#). The central thing to realize here is that #8=2^3#; which means #xlog8=xlog2^3#. And again using the property described above, #xlog2^3=3xlog2#.
We have:
#p=xlog2^3=3xlog2#
Expressing #xlog2# in terms of #p# is now drastically easier. If we take the equation #p=3xlog2# and divide it by #3#, we get:
#p/3=xlog2#
And voila - we have expressed #xlog2# in terms of #p#.
