# 2^x=12+4x Solve the equation

1. Aug 27, 2015

### WeiLoong

1. The problem statement, all variables and given/known data

2^x=12+4x Find x.
2. Relevant equations

none
3. The attempt at a solution
I have tried to change it in form of logarithm xlog2=log4+log3+x and i found that log(3+x) is unexpandable so i was stuck here.

2. Aug 27, 2015

### BvU

Hello Weiloong, welcome to PF !

Your change isn't correctly written: it is true that log(2x) = x log 2, but for log (12 + 4x) you meant log 4 + log(3+x) and that indeed doesn't help all that much.

Why don't you do some trial and error and see what comes out: x = 1, 2, 3, etc...

3. Aug 27, 2015

### Geofleur

You could also try plotting the left and right sides of the equation and seeing where the curves meet, though that only gives an approximate answer.

4. Aug 27, 2015

### Staff: Mentor

Isn't the Quadratic Equation relevant here?

5. Aug 27, 2015

### SteamKing

Staff Emeritus
I don't think so, since the LHS has 2x rather than x2.

6. Aug 27, 2015

### Staff: Mentor

7. Aug 29, 2015

### HallsofIvy

That looks like a candidate for the Lambert W function.

8. Aug 29, 2015

### ehild

You can rewrite the equation as x=2(x-2)-3. Both sides are increasing functions. Plotting them, you can see where the roots are. For positive x, it should be greater than 4. If x is negative, it should be near -3. Trial and error will work :)