1. Sep 20, 2015

### Kevin Licer

So, recently I have been learning about logarithms and how to solve exponential equations with the help of logarithms, but I am curious if I can take the log of both sides of an equation like this in order to solve it?

If not, then could someone explain why? Thanks.

2. Sep 20, 2015

### Orodruin

Staff Emeritus
You cannot, the logarithm of a sum is not equal to the sum of the logarithms of the individual terms.

3. Sep 20, 2015

### Kevin Licer

So, taking the log of both sides of this equation would be more like this?

And would it be useful in this case?

4. Sep 20, 2015

### Orodruin

Staff Emeritus
Yes.

5. Sep 20, 2015

### Mentallic

Yes, that would be the correct way to take the log of both sides, but since you can't do anything with that log on the left because of all the summed terms in its argument, you similarly are stuck at solving this problem. It's a very difficult one to solve that's outside of the scope of what you're doing. You can get lucky and find an easy interger solution though (although this isn't the only solution and you won't be finding the others).

6. Sep 20, 2015

### Staff: Mentor

Orodruin's answer is a response to the first question, I'm pretty sure. As for the second question, would it be useful?
No, for the reason already given, that the log of a sum can't be simplified further.

7. Sep 20, 2015

### Kevin Licer

Yes, I see now, I thought I could solve the equation by taking the log of both sides, but seeing as that is a mistake I'll try and work it out another way. Thanks!

8. Sep 20, 2015

### Curious3141

The most common way of solving a problem like this in an elementary fashion is to try to get it into the form of a polynomial equation that can be easily solved. Of course, the question has to be set specifically so that it's "easy" to solve that way, and in this case, it is.

You need to be familiar with the rules of exponentiation. These are what you need: $a^{b+c} = a^b \cdot a^c$ and $a^{bc} = (a^b)^c$

Using those rules, you can convert your equation into a sextic (power-6) equation in terms of a new variable $y$, where $y = 2^x$. This equation (thankfully) has at least one nice integer solution, which you can quickly find by trial and error (aided by the Rational Root Theorem, http://www.purplemath.com/modules/rtnlroot.htm).

At this point you need to find the full solution set, or at least show there are no other real, positive solutions (since an exponential is strictly positive). You can do this by either polynomial long division or the much simpler synthetic division (http://www.purplemath.com/modules/synthdiv.htm) by the linear factor you uncovered in the last step to get a quintic (power-5) polynomial. It's "tough" to solve or even sketch this curve, but fortunately, there is a little trick called the Decartes' Rule of Signs (http://www.purplemath.com/modules/drofsign.htm) that allows you to very quickly conclude that the quintic has no positive real roots. Hence the single value for $y$ you've already found is the only one that is valid.

Finally, you find $x$ by solving $2^x = y$ for $x$, which I assume you know to do. In the general case you can do so by taking logs of both sides, but here all that's required is simple inspection (this should give you a big hint about how "nice" the solution is).

(Edited for more details, since I was rushing to work in the morning and was unable to complete the post).

Last edited: Sep 20, 2015