Question

How do you solve `1/2=10^(1.5x)`?

Answer 1

x = -0.200686663

Explanation:

`1 / 2 = 10^(1.5 * x)`
Take Log on both sides
`log (1 / 2) = 1.5 x * log 10`
(Hope you are aware of the relationship
`log a^x = x * log a`)
Now `log ( 1 / 2) = log 2^-1 = -1 * log 2`
log 2 = 0.3010300 (this you can get from any log tables or calculator)
So, `-0.3010300 = 1.5 x`
or `x = -0.3010300 / 1.5 = -0.200686663`
Hope it is clear

Answer 2

`:.x=-2/3log_10 2`

Explanation:

tking `log_10` on both sides

`log_10(1/2)=log_10 10^(1.5x)`

`=>log_10 1-log_10 2=log_10 10^(1.5x)`

`=>0-log_10 2=x*1.5xxlog_10 10`

`=>0-log_10 2=(3x)/2`
`:.x=-2/3log_10 2`