Question

How do you solve `2^(x+1)=3^x`?

Answer 1

I would write it as:
`2^x*2^1=3^x`
`(2^x)/(3^x)=1/2`
`(2/3)^x=1/2`
take the log of both sides:
`ln(2/3)^x=ln(1/2)`
`xln(2/3)=ln(1/2)`
`x=1.71`

Answer 2

The answer is: `x=ln2/(ln3-ln2)`.

`2^(x+1)=3^xrArrln2^(x+1)=ln3^xrArr(x+1)ln2=xln3rArr`

`xln2+ln2=xln3rArrx(ln2-ln3)=-ln2rArrx=-ln2/(ln2-ln3)=ln2/(ln3-ln2)`.