Exp. Equations w/ common base - Help!

Azalin

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Jun 8, 2015
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Hey guys. I was pretty sure I had this down but clearly I messed something up. Here's what I have:

3^(n+4)=27^(2n)

Solving for n:

3^(n+4)=(3^3)^(2n)

3^(n+4)=3^6n

n+4=6n

4=5n

4/5=n

I was so sure this was right but my calculator says it's false. Please tell me where I went wrong!

Thanks guys.
 
Hey guys. I was pretty sure I had this down but clearly I messed something up. Here's what I have:

3^(n+4)=27^(2n)

Solving for n:

3^(n+4)=(3^3)^(2n)

3^(n+4)=3^6n

n+4=6n

4=5n

4/5=n

I was so sure this was right but my calculator says it's false. Please tell me where I went wrong!

Thanks guys.

As you have posted the problem - your solution is correct.

What exactly is your calculator "displaying" to make you think that your calculated answer is incorrect?

There is a complex solution to the problem (since fractional powers are involved)

n = [4*log(3) + 2*i*π*m] / [5*log(3)], m element of Z

The answer you calculated is the Real part of the equation above (or the solution at m=0)
 
Hey guys. I was pretty sure I had this down but clearly I messed something up. Here's what I have:

3^(n+4)=27^(2n)

Solving for n:

3^(n+4)=(3^3)^(2n)

3^(n+4)=3^6n

n+4=6n

4=5n

4/5=n

I was so sure this was right but my calculator says it's false. Please tell me where I went wrong!

Thanks guys.

Let's see if you are correct (although you were already told that you were correct).

3^(4/5 +4) = 3^(24/5) since 4/5 + 4 = 4 4/5 = 24/5

Now 27^(2*4/5)= 27^(8/5) = (3^3)^(8/5) = 3^(3*8/5)=3^(24/5)
Looks like it works to me! Good job.

You are more than welcome to post here whenever you like but to be a better student why not try to see for yourself if you are correct or not (and stop using that calculator!!)
 
Last edited:
Well... that's great, thank you very much. Here's what I put into my calculator:


3^((4/5)+4) = 27^(2(4/5))

Which comes back as false. However if I put in each expression individually they return equivalent answers. So maybe I'm inputting it incorrectly as a whole equation.
 
Let's see if you are correct (although you were already told that you were correct).

3^(4/5 +4) = 3^(24/5) since 4/5 + 4 = 4 4/5 = 24/5

Now 27^(2*4/5)= 27^(8/5) = (3^3)^(8/5) = 3^(3*8/5)=3^(24/5)
Looks like it works to me! Good job.

You are more than welcome to post here whenever you like but to be a better student why not try to see for yourself if you are correct or not (and stop using that calculator!!)

I totally agree and I generally try to only use the calculator to check my answers. That's why this one had me stumped so bad and made me question my answer :)

Thanks for the confirmation!
 
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