Exponential Equation: how to find the solution of this equation? 2^(1-x) = 3

[Illvoices]

So how to find the solution of this equation?
2^1-x=3

what i did was add log to both sides of the equation since in the example they add log. Then i divided 3log and 2log, then i added 1 and got 2.5849625007211561814537389439478

 

[JeffM]

So how to find the solution of this equation?
2^1-x=3

what i did was add log to both sides of the equation since in the example they add log. Then i divided 3log and 2log by what then i added 1 why and got 2.5849625007211561814537389439478

What you mean is that you TOOK the log of both sides of the equation. Question: why can you legitimately do that?

\(\displaystyle 2^{(1 - x)} = 3 \implies log \left ( 2^{(1 - x)} \right ) = log(3) \implies what?\)

 

[HallsofIvy]

As JeffM said, you took the logarithm of both sides (mathematically, you cannot 'add' log to both sides because log is not a number). Okay so you got \(\displaystyle log(2^{1- x})= (1- x) log(2)= log(3)\). Then you did not "divide 3log and 2log", you divided both sides by log(2) to get \(\displaystyle 1- x= \frac{log(3)}{log(2)}\).

But "adding 1 to both sides of that gives 2- x, not x! If instead you subtract 1, you get "-x= a number".