Question: If`log_8 x + log_8 1/6 = 1/3,` then the value of x is :
Note: `log x + log_8 1/6`
`= 1/3 ⇔ (1og x)/(log 8) + (log 1/6)/(log 8)`
`= 1/3`
`:. log x + log 1/6`
`= 1/3 log 8`
`or log x + log 1/6`
`= log (8^1/3)`
`= log (2^3)^1/3.`
`:. log x + log 1/6`
`= log 2`
`or log x`
`= log 2 - log 1/6`
`= log (2 xx 6/1)`
`= log 12.`
`:. x = 12.`