help my friend with his math

[ButterflyBlood]

ok, each of these logs is base 3:

log(x+3) + log(x-3) = 2

can anyone figure out x?

[TXlib]

1. here

log(x+3) + log(x-3) = log{(x+3)(x-3)} = 2

I assume that's log base 10?

then exponentiate both sides:

10^log{(x+3)(x-3)} = (x+3)(x-3) = 10^2 = 100

x^2 - 9 = 100

x^2 = 109

x = +/- sqrt(109)

---

On Edit:
Whoops, I didn't read your message closely enough...

Use base 3 to exponentiate both sides:

3^log{(x+3)(x-3)} = (x+3)(x-3) = 3^2 = 9

x^2 - 9 = 9

x^2 = 18

x = +/- sqrt(18)

[Waistdeep]

9. No minus, just plus

In this problem the negative root would be disallowed because both (x-3) and (x+3) would be negative and the log of a negative number gets you into complex numbers, which I'm sure wasn't intended in this problem.

So the answer is just x = +sqrt(18) = +3 sqrt(2)

[TXlib]

18. D'oh! You're right!

Ahhh, the dangers of mixing tequila and mathematics...

Don't drink and derive.

[onebigbadwulf]

2. Yes

Okay to add logs you multiply their components

log (a) + log (b) = log (ab)


so log3 (x^2 - 9) = 2


And because we have to work in log 10 we must use the rule


log b (x) = log 10 (x)/ log10 (b)


soooo


log10 (x^2-9)/ log10 (3) = 2



multiply over

log10 (x^2 - 9) = 2 log10 (3)

///////////////////////editted back to original///////

log10 (x^2 - 9) = log10 (3^2)

so x^2 - 9 = 9
x^2 = 18

or x= sqrt 18


GOT IT

[TXlib]

6. You were right up til log10 (x^2 - 9) = 2 log10 (3)

log10 (x^2 - 9) = 2 log10 (3)



log10 (x^2 - 9) = log10 (3^(1+2))

so x^2 - 9 = 27

x^2 = 36

or x= 6

a logb(c) = logb(c^a), where logb(x) is log base b of x.

so, 2 log10 (3) = log10 (3^2), not log10 (3^(1+2)).

[question everything]

7. I think that you are mistaken

Let's go to the line where you have everything on the basis of 10:

log10(x^2-9) = 2 log10 (3)

Next should be

log10 (x^2-9) = log10 (3^2), not 3(1+2) as you wrote

thus

log10 (x^2-9) = log10 (9)

x^2 - 9 = 9

x^2 = 18

x = sq root of 18

You can reach the same conclusion without going through the conversion to the basis of 10:

log3(x^2-9) = 2

(x^2-9) = 3^2

x^2 - 9 = 9

x^2 = 18

x = sq root of 18

[onebigbadwulf]

8. No because

2 log (x) = log (x)^2


and the law of exponents says that bringing an exponent inside brackets means you add them.


so


log (x)^2 = (x^1)^2 = (x^(1+2)) = (x^3)



I roxor

[TXlib]

10. OK, let's try your way out with numbers

Assume base 10 to make it easy:

5 log 100 = 5 * 2 = 10

According to you, 5 log 100 = (log 100)^5 = 2^5 = 32, if I'm interpreting your brackets right.

The law of exponents says:

a log(b^c) = log(b^(a*c))

not log(b^(a+c)).

[Waistdeep]

11. Onebigbadwulf, you're definitely wrong

Substitute your x=6 into the original equation

log(x+3) + log(x-3) = 2
log(6+3) + log(6-3) = 2
log(9) + log(3) = 2

since we're working in base 3, log(9) = 2, log(3) = 1

2 + 1 = 2 ??????


sqrt(18) = 3 sqrt(2) is correct.

[onebigbadwulf]

13. damn

that was my original answer too but it didnt look right

[question everything]

12. You are wrong

You do not bring an exponent inside a bracket

You X is not an exponent.


thus, the sample above is:

2 log X = log X^2

[TXlib]

14. You get 9/10 points

if x^2 = 18, then
x = +/- sqrt(18)

Don't forget the other root!

(Former physics teacher here...)

[Waistdeep]

15. Other root doesn't work

See post 9., above

Former physics teacher here, too.

[question everything]

17. I don't even care to remember of how many years it has been

since I solved such problems, or just used logs in my work ;-)

[onebigbadwulf]

3. I am a genius

I fixed my post now it's flawless

[Parrcrow]

4. lets see

DON'T PANIC

the answer is 42. The answer to everything is 42.

[DiverDave]

16. Yep 42!

why go through all that?...42

Dont Panic

[LiberalVoice]

5. All these numbers are scarying me!

Make it stop mommy! Make it stop!