Question

How do you solve `2logx=1.6`?

Answer 1

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`