# Solving Exponential Equations Using Logarithms

1. Jan 16, 2004

### JDK

Hello,

The following problem is bothering me quite a bit. It is...

Solve.
3 x 2^x = 12

The Unit this question is in is about solving exponential equations through expressing each side as a power of the same base, by taking a base (x) logarithm of each side or taking a base 10 logarithm of each side.

The answer is supposed to be 2. Here are my calculations.

3 x 2^x = 12
log 3 + xlog 2 = log 12
x(log 3 - log 2) = log 12
xlog 3/2 = log 12
x = log 12 / log 2/3

Now this is incorrect for probably more then one reason. I've had success doing the questions of the form... 3^3x-1 = 9^2x. But when they introduced multiplication on the left side of the eq I got lost. Also I noticed that 3 or 2 cannot have a whole number exponent to make 12. Confusion is upon me. Thanks for the help in advance!

2. Jan 16, 2004

### himanshu121

How u reached x(log 3 - log 2) = log 12 from
log 3 + xlog 2 = log 12

3. Jan 16, 2004

### JDK

I was trying to work along with the text books' example but the question didn't follow mine very closely.

4. Jan 16, 2004

### Muzza

As was stated, this is where you go wrong:

You can not factor x out of the log(3) term... since it has no factor of x in it! And what's up with the minus sign?

The correct way to do it is something like this:

log(3) + xlog(2) = log(12)
xlog(2) = log(12) - log(3)
xlog(2) = log(12/3)
x = log(4)/log(2) = log(2^2)/log(2) = 2log(2)/log(2) = 2

However, a simpler/better way is this:

3 * 2^x = 12
2^x = 12/3 = 4
2^x = 2^2
x = 2

:P

Last edited: Jan 16, 2004
5. Jan 16, 2004

### Warr

the main thing you should focus on is getting x by itself, right off

so from $$(3)2^x = 12$$, you should bring the 3 over to the right side, and then use logarithms to separate the 2 and the x. Then you can bring the 2 over, isolating x. Also remember that $$log (X^y) = (y)(log X)$$, although I think you already know this from the stuff you did. Try this and if you still can't get it post back.

Last edited: Jan 16, 2004
6. Jan 16, 2004

### HallsofIvy

One unfortunate complication is that you used "x" to mean two different things. My first reaction when I saw "3 x 2^x = 12" is that you can't do this using logarithms since you have x both in the exponent and not in the exponent! Then I realized that the first "x" is times. It would be better to write the equation either as
"3*2^x= 12" using "*" for times or to just write 3(2^x)= 12.

As has been pointed out, you are correct that
log(3)+ x log(2)= log 12. Your error is in then writing
"x(log 3 - log 2)= log 12". That would be the same as
"xlog(3)+ xlog(2)= log 12". What you should do from
log(3)+ x log(2)= log 12 is subtract log(3) from both sides:
x log(2)= log(12)- log(3)= log(12/3)= log(4).

Actually, the way I would be inclined to do the problem is to divide by the "3" first: Dividing both sides of 3(2^x)= 12 gives 2^x= 4. You should then be able to recognize that 2^2= 4 so x= 2. That is, in a sense, using logarithms because it uses the basic definition of logarithm. Given that a^x= a^b, it follows that x= b because we know that a^x is a one-to-one function: it HAS an inverse. Saying that x= b is the same as using the inverse on both sides of the equation and the inverse IS the logarithm.

7. Jan 16, 2004

### Warr

however, he will not always be able to come to those conclusions because x will not always be that simple. I suggest that instead of the last steps being:

2^x = 4
2^x = 2^2

$$log 2^x = log 4$$
$$x(log 2) = log 4$$
$$x = \frac{\log 2}{log 4}$$

8. Jan 16, 2004

### himanshu121

it should be
$$x = \frac{\log 4}{log 2}$$

9. Jan 17, 2004

### Warr

ack, sorry, the latex code was annoying me a bit and I must have accidently switched them

10. Jan 17, 2004

### HallsofIvy

Yes, it is true that if the problem were for example, 3*2x= 9, then it reduces to 2x= 3 and so x log(2)= log(3) so x= log(3)/log(2).

However, in my opinion, it is much better to understand that logarithm means the exponent. I suspect that the whole point of the problem 3(2x)= 12 was to recognize that this is the same as 2x= 22 rather than to mindlessly apply a formula.

11. Jan 19, 2004

### JDK

Wow. I actually understand this now. You all have my many thanks and my praise. I can continue with my work finally. This is the best response for help I've recieved yet.