Question

How do you solve `2.4^(3x+1)= 9`?

Answer 1

`x = (log9-log8)/(6log2)=0,02832`

Explanation:

Using laws of exponents and indices you may write the equation as
`2^1*2^(6x)*2^2=3^2`
`therefore2^(6x+3)=3^2`
`therefore2^(6x)= (3^2)/(2^3)=9/8`
Now taking the logarithm on both sides and sing laws of logs we get
`6xlog2=log9-log8`
`therefore x = (log9-log8)/(6log2)=0,02832`