Question

How do you solve `log2sqrt(5x+1) = 4`?

Answer 1

I found:
`x=5xx10^6` if log in base 10
or
`x=51` if log in base 2

Explanation:

Assuming a log in base `10` we can write:
`log_(10)2sqrt(5x+1)=4`

using the definition of log:
`2sqrt(5x+1)=10^4`
`cancel(2)sqrt(5x+1)=10000/2`
`sqrt(5x+1)=5000`

squaring both sides:
`5x+1=(5000)^2`
`x=((5000)^2-1)/5~~5xx10^6`

If on the other hand the base is `2` we can write:
`log_2sqrt(5x+1)=4`
again using the definition of log we get:
`sqrt(5x+1)=2^4`
`sqrt(5x+1)=16`
squaring both sides:
`5x+1=256`
`5x=255`
`x=51`