`log_2(x - 1) = 5` Solve the equation accurate to three decimal places

`log_2 (x-1) =5`

To solve, convert the equation to exponential form.

Take note that if a logarithmic equation is in the form

`y = log_b (x)`

its equivalent exponential equation is

`x= b^y`

So converting 

`log_2 (x-1)=5`

to exponential equation, it becomes

`x-1 = 2^5`

Then, simplify the right side.

`x-1=32`

And isolate the x.

`x= 32 +1`  

`x=33`

Therefore, the solution is `x=33` .