Expert: Sherry Wallin - 11/14/2011

Question
QUESTION: sir how can we solve 2^x+x-3=0,and that type of questions ,please give me the method .

ANSWER: Owais~

2^x+x-3 = 0 by inspection you can see that x = 1 is a solution: 2^1 + 1 - 3 = 2 + 1 - 3 = 3-3 = 0
You can also graph the exponential equation and see where it crosses the x axis and you will find that it crosses at (1,0) also.

Math Prof

---------- FOLLOW-UP ----------

QUESTION: sir but any other way like doing some algebra can solve this problem or any oter way accept graphing and judging?

Answer
There are many ways to try to bound the problem but I do not know your level of math. Have you had calculus?
If you take the ln of both sides you get ln(2^x+x-3) = ln 0 but you can't do that because log functions are not defined at 0 and less. So move the -3 to the right and get
ln(2^x+x) = ln 3 = 1.098612289 but now how would you solve for x? If you move x to the other side initially you would get 2^x = 3-x and now take the ln of both sides getting
ln(2^x) = ln(3-x). You always have the option of taking the log base x of both sides providing you restrict x to be greater than 0:
log_x(2^x) = log_x(3-x) => 2 = log_x(3-x)
Note you can't get x entirely by itself...but you can bound it using real analysis.


2^x = 3 - x  take the log_2 of both sides
log_2(2^x) = log_2(3-x)
x = log_2(3-x)  
Notice that 3-x > 0 => x < 3 but x > 0 so 0 < x < 3
You have bounded what x is and by inspection you can see that x is 1. As I noted earlier log functions log x pass through (1,0) since log_b(x) = y => b^y = x and that only happens at
y = 0 and x = 1

There are many ways to try to solve the problem but I do not know your level of math. Have you had calculus?
If you take the ln of both sides you get ln(2^x+x-3) = ln 0 but you can't do that because log functions are not defined at 0. So move the -3 to the right and get
ln(2^x+x) = ln 3 = 1.098612289 but now how would you solve for x? If you move x to the other side initially you would get 2^x = 3-x and now take the ln of both sides getting
ln(2^x) = ln(3-x). You always have the option of taking the log base x of both sides providing you restrict x to be greater than 0:
log_x(2^x) = log_x(3-x) => 2 = log_x(3-x)
Note you can't get x entirely by itself...but you can bound it (real analysis)


2^x = 3 - x  take the log_2 of both sides
log_2(2^x) = log_2(3-x)
x = log_2(3-x)  
Notice that 3-x > 0 => x < 3 but x > 0 so 0 < x < 3
You have bounded what x is and by inspection you can see that x is 1. As I noted earlier log functions log x pass through (1,0) since log_b(x) = y => b^y = x and that only happens at
y = 0 and x = 1

Math Prof


Math Prof