How do you solve #2logx=1.6#?

1 Answer
May 26, 2016

Answer:

i found: #x=b^0.8# where #b# is the base of your log.

Explanation:

It depends on the log's base but we can say that is, say, #b#, so we have:
#2log_b(x)=1.6#
rearranging:
#2log_b(x)=16/10#
#log_b(x)=1/2*16/10#
#log_b(x)=1/cancel(2)*cancel(16)^8/10=0.8#

using the definition of log:
#x=b^0.8#

Now you can use the appropriate #b# for our case:

if #b=10#, for example:

#x=10^0.8=6.30957#