Question 287532
<pre>
Solve Log2(x+1)=5
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The other person who responded has the following, which makes
ABSOLUTELY no SENSE, in this author's opinion!
"log ( 2x + 2 ) = 5
=> e^5 = 2x + 2
=> 2x = e^5 -2
Now do the rest....
Happy"
{{{log (2, (x + 1)) = 5}}}, with x > - 1
          {{{x + 1 = 2^5}}} ------ Converting to EXPONENTIAL form
         x + 1 = 32              
               <font color = red><font size = 3>x</font></font> = 32 - 1 = <font color = red><font size = 3>31</font></font>

That's IT!!</font></font></font></b></pre>