How Do You Solve These Exponential Equations?

[JBD2]

Homework Statement

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}

Homework Equations

None that I am aware of.

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 textbook is 7.

Answer Key:
Question 1:
\frac{3}{5}
Question 2:
7

Thanks to everyone who helps.

 

[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.

 

[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

 

[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.

 

[HallsofIvy]

Homework Statement

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}

Homework Equations

None that I am aware of.

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

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 textbook is 7.

Answer Key:
Question 1:
\frac{3}{5}
Question 2:
7

Thanks to everyone who helps.

 

[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...

 

[HallsofIvy]

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

 

[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.

 

[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.

 

[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?

 

[nicksauce]

<br /> 2^{x}\times2^{-1}-2^{x}=2^{-3}<br />

You can write this as
<br /> 2^x(2^{-1}-1})=2^{-3}<br />

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

 

[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)

 

[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)

 

[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?

 

[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.

 

[HallsofIvy]

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

And 3+1= ?

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

 

[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 wouldn't 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?

 

[nicksauce]

Yes that makes sense.

 

[JBD2]

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

 

[cheff3r]

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.

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

 

[HallsofIvy]

A fairly obvious way to solve 3^x(4)= 36 is to first divide both sides by 4: 3^x= 9.

 

[JBD2]

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

That is equal to 27 and not 36 (3 x 9) so it doesn't work out.

A fairly obvious way to solve 3^x(4)= 36 is to first divide both sides by 4: 3^x= 9.

Oh ok thanks that makes sense.