# Homework Help: Isolating a function variable

1. Sep 30, 2012

### Tyrone123

1. The problem statement, all variables and given/known data

Im supposed to calculate the value of x for

20*log10($\frac{\sqrt{x^2 +1}}{cos(x)}$) = 3

3. The attempt at a solution
I start off by dividing by 20 on both sides so I just have the expression with log.. and my idea henceforth is to take the inverse of cosine to get rid of cos(x), and square the square root to cancel out.. but the logarithm part has me stuck.. and im not sure how to get rid of/move it.. ive thought about multiplying with its inverse, but that has not gotten me any results.. any tips?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Sep 30, 2012
2. Sep 30, 2012

### Robert1986

How about $e$?

But I can't really tell what to do after that.

3. Sep 30, 2012

### SammyS

Staff Emeritus

Now, there is no equation!

(Formerly it said 20*log$\frac{\sqrt{x^2 +1}}{\cos(x)}=3\ .$)

4. Sep 30, 2012

### Tyrone123

woops, I edited log to log10.. and suddenly the equation was gone :b

5. Sep 30, 2012

### ohms law

I think propertieds of logs are important here.

log10 u/v = log10(u)-log10(v)

So, your original equation can be rewritten as:
20(log10(u)-log10(v))=3
with the numerator and denominator substituted for u and v, respectively. (hopefuly that makes sense)

Does that help?

6. Sep 30, 2012

### Tyrone123

Possibly, ill try

7. Sep 30, 2012

### Tyrone123

Not really.. I end up with a slightly prettier equation, but im not sure how to isolate x from log(cos(x)

8. Sep 30, 2012

### ohms law

ah, well, there's a trig identity to help with that, I'm sure...
Off of the top of my head, there's always cos^2(x)+sin^2(x)=1, so...
I'll have to look some stuff up and see if it'll help.

9. Sep 30, 2012

### SammyS

Staff Emeritus
If $\log_{10}(A)=B\,,$ then $A=10^B\ .$

10. Sep 30, 2012

### Robert1986

All I see right now is something like this (where I'm using 10 instead of e) $\sqrt{x^2 + 1} = a \cos{x}$ where $a = 10^{3/20}$ then use the trig identity ohms law mentioned. (But multiply the identity by -1)

11. Sep 30, 2012

### Tyrone123

Hmm.. so far I have at least gotten rid of the log part, and I now have:

$\frac{\sqrt{x^2 +1}}{cos(x)}$ = 10$\frac{3}{20}$

Which I can rearrange a bit.. but that cos(x) is causing me troubles!

12. Sep 30, 2012

### ohms law

fyi, for the numerator portion, I simplified it to 1/3 log(x)

13. Sep 30, 2012

### SammyS

Staff Emeritus
I do believe that solving this requires using a numerical method.

14. Sep 30, 2012

### ohms law

1/cos(x) would = sec(x)...
so, log(1/cos(x)) = log(1) - log(sec(x))
since log(1) = 0, we can just get rid of that

just not sure how much that helps

15. Sep 30, 2012

### ohms law

so, you're saying just plug and puke?
:)

16. Sep 30, 2012

### SammyS

Staff Emeritus
I've reviewed my posts, and no, I didn't say that. (LOL !)

OP may want to put his equation into a more suitable form before trying a numerical method.

Also, a graph may help in choosing a starting approximation.

Last edited: Sep 30, 2012
17. Sep 30, 2012

### ohms law

I missed this earlier, but... I'm not sure what you mean with that equality. It's not true as stated (see graph):

18. Sep 30, 2012

### ohms law

I guess that I'm just not clear what you mean by "requires using a numerical method".
I feel like I should know what you mean, but...

19. Sep 30, 2012

### Tyrone123

I dont mean anything in particular, just me trying to isolate x in the equation

20. Sep 30, 2012

### SammyS

Staff Emeritus
Actually, that equality is true !

It leads to $\displaystyle \sqrt{x^2+1}=10^{3/20}\cos(x)\ .$

Tyrone123,
Notice that $\left(10^{3/20}\right)^2\approx1.99526\,,$ which is very nearly 2, so that $10^{3/20}$ is just a little less that $\sqrt{2}\ .$

Also, it may help to square that above equation, which then becomes:
$\displaystyle x^2 +1=10^{3/10}\cos^2(x)\ .$​

For small values of x, a good approximation for cos(x) is:
$\displaystyle \cos(x)\approx 1-\frac{x^2}{2}\ .$​

Plug that into the previous equation and solve for x to get a good starting place for any numerical method you choose.

21. Sep 30, 2012

### ohms law

man, trig makes my head hurt... :grumpy:
I don't get it. I'd drop it, but my sense is that this is part of the problem that Tyrone is having too (maybe I'm wrong, but...).

I know that graphs don't prove anything, but the graphs are (extremely) different (even with the cos(x) product added to the one side), so... I'm confused.

22. Sep 30, 2012

### SammyS

Staff Emeritus
Here are the graphs of $y=\sqrt{x^2+1}$ and $y=10^{3/20}\cos(x)$ overlaid on the same graph (from WolframAlpha).

#### Attached Files:

• ###### Graphical_solution.gif
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23. Sep 30, 2012

### SammyS

Staff Emeritus
What is graphed here?

24. Sep 30, 2012

### SammyS

Staff Emeritus
Take Tyrone's original equation.
$\displaystyle 20\,\log_{10}\left(\frac{\sqrt{x^2 +1}}{\cos(x)}\right)=3\$​

Do a little basic algebra & apply some of the log rules you suggested:
$\displaystyle \log(\sqrt{x^2+1})-\log(\cos(x))=\frac{3}{20}$

$\displaystyle \log(\sqrt{x^2+1})=\frac{3}{20}+\log(\cos(x))$

If you use each side as the exponent with a base of 10, you get:
$\displaystyle \sqrt{x^2+1}=10^{3/20}\cos(x)\$​