# Solving Exponential Equations

1. Aug 21, 2008

### JBD2

1. The problem statement, all variables and given/known data
Solve the following:
Question 1:

$$\sqrt[5]{256}\div\sqrt[6]{64}=2^{x}$$

Question 2:

$$\frac{(9^{2x-1})^{3}(3^{3x})^{2}}{(27^{x+2})^{4}}=81^{3}$$

2. Relevant equations
None that I am aware of.

3. The attempt at a solution
Question 1:

$$\sqrt[5]{2^{8}}\div2^{1}=2^{x}$$

Question 2:

$$\frac{(3^{4x-2})^{3}(3^{6x})}{(3^{3x+6})^{4}}=3^{12}$$

$$3^{12x-6+6x-(12x+24)}=3^{12}$$

$$6x=39$$

$$x=6.5$$

Sorry that the attempt for question 1 was lacking so much, I was just unsure of what to do with the square roots etc...As for number 2, the answer I got is incorrect, when it says that the answer in the back of my text book is 7.

Question 1:
$$\frac{3}{5}$$
Question 2:
7

Thanks to everyone who helps.

2. Aug 21, 2008

### nicksauce

For #1 you almost have it... just express the fifth root of 2^8 as 2 raised to some power, then use the fact that 2^a / 2^b = 2^(a-b).

For #2 just check your numbers again. -6 -24 = -30. 12 + 30=42.

3. Aug 22, 2008

### kenewbie

1)

Two things that is helpful to know:

$$\sqrt[a]{n^b} = n^{ \frac{b}{a} }$$

and

$$\lg{\frac{a}{b}} = \lg{a} - lg{b}$$

You don't need the logarithm in this case though, since all your numbers are factors of two.

k

4. Aug 22, 2008

### JBD2

Ok thanks, I realized what I had done wrong for number two earlier today but I didn't know the rule about number one. I'm just starting into the logarithms so things should start becoming more familiar as well. Thank you.

5. Aug 23, 2008

### HallsofIvy

Staff Emeritus
Well, how about these:
$$^n\sqrt{a}= a^\frac{1}{n}$$
$$(a^n)^m= a^{mn}$$
$$a^m\div a^n= a^{m-n}$$

6. Aug 23, 2008

### JBD2

Yes sorry I should've added those (the ones I used). I have another question though, I don't know what to do when adding two variables with exponents, for example what would something like this become:

$$x^{4}+x^{3}=$$

or

$$x^{4}-x^{7}=$$

I couldn't find it on the internet and I don't remember what to do in this case...

7. Aug 23, 2008

### HallsofIvy

Staff Emeritus
There is no rule for adding different powers of the same base. That is exactly why polynomials are written as they are.

8. Aug 23, 2008

### JBD2

Ok I've tried so many things and I can't seem to figure out these two questions. Here they are:

Question 3:

$$2^{x-1}-2^{x}=2^{-3}$$

Question 4:

$$3^{x+1}+3^{x}=36$$

My work is a mess as I have tried so many variations which are probably way off course.

9. Aug 23, 2008

### nicksauce

you can write 2^(x-1) as 2^x*2^(-1)
Thus you can factor the LHS, and isolate 2^x. Question 4 should be similar.

10. Aug 23, 2008

### JBD2

Can you explain what you mean? So it would become:

$$2^{x}\times2^{-1}-2^{x}=2^{-3}$$

But I can't really combine the like terms ($$2^{x}$$), can I? Because isn't the first one considered "attached" to the second one? ($$2^{x} and 2^{-1}$$)

And for the second question:

$$3^{x}\times3^{1}+3^{x}=36$$

$$6^{x}\times3=36$$

$$18^{x}=36$$

I'm not sure what to do from here...maybe I'm already wrong?

11. Aug 24, 2008

### nicksauce

$$2^{x}\times2^{-1}-2^{x}=2^{-3}$$

You can write this as
$$2^x(2^{-1}-1})=2^{-3}$$

Now write the term in the parentheses as 2 to some power, and you should be all set.

12. Aug 24, 2008

### JBD2

So it's:

$$2^{x}(-2^{-1})=2^{3}$$

But seeing as the answer in the back says no solution, I'm assuming it's not possible to solve this with a negative base seeing as the rest are positive? (2 and -2)

13. Aug 24, 2008

### nicksauce

Right. You have 2^x = (some negative number). There is no value of x for which this is true, as 2^x is always positive. (I should have noticed this earlier, but didn't think of it for what ever reason)

14. Aug 24, 2008

### JBD2

Oh ok that makes sense, with the other question (question 4) I know how to go about it now but how do I establish a common base between 3 and 36?

15. Aug 24, 2008

### JBD2

Actually I'll just show my work for now so you can see the progress I've made so far:

$$3^{x+1}+3^{x}=36$$

$$3^{x}(3^{1}+1)=36$$

Ok nevermind I thought I had an idea but I'll leave this up for reference.

16. Aug 24, 2008

### HallsofIvy

Staff Emeritus
And 3+1= ?

17. Aug 24, 2008

### JBD2

$$3^{x}(4)=36$$

Well its obvious that x is 2 but how do I do that algebraically...I can't go $$12^{x}=36$$ because x=2 wouldnt work...This probably looks like such a dumb question I just can't figure out how to get a common base...Can I go:

$$3^{x}(4)=3^{2}(4)$$

$$x=2$$

Does that make sense?

18. Aug 24, 2008

### nicksauce

Yes that makes sense.

19. Aug 24, 2008

### JBD2

Oh ok thank you, I just didn't know you could have a base with two separate numbers.

20. Aug 25, 2008

### cheff3r

leave the left hand side as it was (from the start) change right hand side to
$$3*3^{2}$$ note this is equal to 36
now all the bases are the same and they can cancel/disapear (index laws)
so now you have
(x+1)+x=1+2
and i'm sure you can go from here