1. Dec 7, 2004

### aisha

(3squareroot27)^3x-1=9^x+3
I dont get these questions because I dont know how to get rid or the square root. First I thought Id multiply both sides by the root 3 and then the square root would disappear and i'd be left with (27)^3x-1=(9^x+3)^3 then I found a common base which could be 9? or does it have to be 3? Well I used 9 and when I simplified the exponents I got x=2 but the answer is x=7 can someone help me out plz?

2. Dec 7, 2004

### dextercioby

I can't make head or tail of your equation.You have two choises of making me help u:either u post it using Latex or write it in words if u don't know how to place some paranthesis.
Anyhow,if the equation is:
$$(3\sqrt{27})^{3x-1} =9^{x+3}$$,then the solution is not 7,but 17/11.

3. Dec 7, 2004

### Banana

If your equation is as dexter thinks, then the first thing I would do is take the cube root of 27 and just change that part to a 3. Then, anytime you have a variable in an exponent, you just take the log or ln of each side. It turns it into an easy equation. For instance to solve 3^x = 12, you just make it into x * log 3 = log 12. So to get x, just use your calculator to divide log 12/log 3. You should be able to do your problem this way.

4. Dec 7, 2004

5. Dec 7, 2004

### dextercioby

I wasn't wrong.To that specific equation i wrote that the solution is 17/11.
The equation:
$$(\sqrt[3] {27})^{3x-1} = 9^{x+3}$$
has the obvious solution "x=7".Compare the 2 equations and see the're not identical.
I believe it is this second eq (the one with the solution 7,not the first,with the solution 17/11) that he/she initially wanted to write,but he/she didn't didn't succeed in placing to gd paranthesis and saying that in the first paranthesis (i wonder where did those 2 paranthesis came from...) it was $\sqrt[3] {27}$ and not the $3\sqrt{27}$,as i correctely interpreted.

Ps.I'm neither blind,nor stupid. :grumpy:

6. Dec 7, 2004

### aisha

OH MY, SORRY EVERYONE, I dont know how to use the equation editor and yes the answer is 7 from my text book but the question was cube root of 27
not 3squareroot27 , but none of u helped me to solve these types of problems? I wanted to know how to get rid of the square root and then solve? This year we are not learning log.

7. Dec 7, 2004

### dextercioby

1.Click one on the formulas written in TEX.it will open u a small window in which u'll see the code (the letters,numbers&signs) used to type that formula plus a link to antothe page where u'll have the .pdf latex guide file.Better dld it if u're using your pc to access PF.If after browsing the 4 pages,u don't understand how to edit formulas,then go to another thread on this forum where in about 100 posts u'll be explained how and what to do.I believe it's in the "general physics" forum and it has a title "introducing latex typesetting" or something like that.You'll see it,it's a "sticky".
2.The third order root of 27 is 3,since exponentiaiting and extracting roots are inverse (to each other) operations and 3 to the power of 3 is exactly 27.

Daniel.