Question

Logarithms

Answer 1

lol write the equation well this time :D

Answer 2

\[\log_{\log_{\log_{...}}} (\infty ^{\infty})=?\]

Answer 3

0 I guess....

Answer 4

\[\huge (0.5)^{\log_3\log_\frac{ 1 }{ 5 }\left( x ^{2}-\frac{ 4 }{ 5 } \right)} < 1\]

Answer 5

Sorry @mathslover I just made that up to joke around, I don't think it has an answer lol.

Answer 6

lol... I just guessed :-P

Answer 7

@No.name - You need to solve for x, right?

Answer 8

Yes

Answer 9

I have a hunch that rewriting .5 as 2^(-1) will help. I'm thinking still.

Answer 10

that will help actually

Answer 11

as , \(\log_{5^{-1}} (x^2 - \cfrac{4}{5}) = -1 \log _ {5} (x^2 - \cfrac{4}{5})\)

so, \( (0.5)^{\log_3\log_\cfrac{ 1 }{ 5 }( x ^{2}-\cfrac{ 4 }{ 5 } )} < 1\) becomes :

\((2) ^ {\log_3 \log_5 (x^2 - \cfrac{4}{5})} < 1\)

Answer 12

We can also write it as :

\((2) ^ {\log_3 \log_5 (x^2 - \cfrac{4}{5})} < 2^{0}\)

Answer 13

then it would be solvable yes

Answer 14

\[.5^{\log_3(\log_{1/5}(x^2-4/5))}<1\]log base 1/2 to both sides\[\log_3(\log_{1/5}(x^2-4/5))<0\] Make both exponents of 3\[\log_{1/5}(x^2-4/5)<1\]Raise these as exponents of 1/5\[x^2-4/5<1/5\]Add 4/5 to both sides\[x^2<1\]x=1

Answer 15

x=-1* whoops

Answer 16

That is what I got..

Answer 17

yes , thanks

Answer 18

Turns out the whole 2^(-1) thing really didn't matter lol.

Answer 19

@Kainui , when you take log base 1/2 in RHS, then shouldn't it be -1 (RHS)?

Answer 20

No, I don't think so, but I'm not sure if I know what you mean.

Answer 21

(1/2)^0 = 1

Answer 22

looking back, I'm pretty sure this is a wrong statement: \[\log_{1/5}a=-\log_5a\] The rule I think you're thinking of is some combination of \[\log_ba=(\log_ca)/(\log_cb)\] and \[-\log(a)=\log(\frac{1}{a})\]

Answer 23

Sorry, I got it...

Answer 24

All good haha no worries. =P

Answer 25

(1/2)^0 =1

Answer 26

but yeah,

\(\log _{5^{-1}} (a) = -1 \log_{5} (a) \)

Answer 27

That's what he means to say , so what was @mathslover 's problem , i didn't get

Answer 28

I used base power rule...

\(\log_{a^m} x = \cfrac{1}{m} \log_a x\)

Answer 29

No, you're saying: \[\log_{1/5}(x)=-\log_5(x)\] but what I'm saying is \[\log_{1/5}(x)=-\log_{1/5}(\frac{1}{x})\]

Answer 30

I'm pretty sure what you're saying isn't a rule, or maybe I'm tired... @_@

Answer 31

It is a rule

Answer 32

Yeah, now I'm pretty sure my answer is wrong after plugging it in. -1 doesn't solve it.

Answer 33

You're also right @Kainui with your second property.

But, the first one you mentioned is actually a rule,

one can use it directly

like if you're supposed to find : \(\log_4 {16} = ?\)

there will be 2 suitable methods :

i) \(\log_4 {4^2} = 2 \log_4{4} = 2\)

ii) \(\log_{2^2} {2^4} = \cfrac{1}{2} \log_2 {2^4} = 2 \log_2 2 = 2 \)

Both works...

Answer 34

Same thing written in two different forms

Answer 35

Well, if I solve it in my way...

I get \(x^2 < \cfrac{29}{5}\)

Answer 36

\(2 ^{\log_3 \log_5 (x^2 - \cfrac{4}{5})} < 2^ 0\)

Thus,

\(\log_3 \log_5 ( x^2 - \cfrac{4}{5} ) < 0\)

\(\log_5 ( x^2 - \cfrac{4}{5}) < 1\)

\(x^2 - \cfrac{4}{5} < 5 \\
x^2 < \cfrac{29}{5} \)

Answer 37

isn't x^2 < 1
mean x<1 and x >-1

Answer 38

Wow it's actually pretty easy to derive from the rules I put out there lol. I feel silly, I've never seen that rule ever before, weird. \[\log_{a^m}(x)=\frac{\log_{a}(x)}{\log_{a}(a^m)}=\frac{1}{m}\log_a(x)\]

@hartnn Yeah, you're obviously completely right. I shouldn't be here lol.

Answer 39

O.O then why am I getting a weird answer? will anyone help me to recognize the mistake in my soln?

Answer 40

I think your solution might be more right than mine is. I'm rethinking it, this is really weird hahahaha.

Answer 41

:-)

Answer 42

Try plugging that in @mathslover and see if you get a correct answer out of it.

Answer 43

k , will try that

Answer 44

http://www.wolframalpha.com/input/?i=2%5E%5Blog_3+%7Blog_5+%2829%2F5-+4%2F5%29%7D%5D+%3D+

Answer 45

I tried for 29/5

that means, if I put any value *less than 29/5* then it will give result which will be \(\bf{<1}\) .

Answer 46

that makes me little bit happy , I finally solved a logarithm question ...

http://www.wolframalpha.com/input/?i=2%5E%5Blog_3+%7Blog_5+%285-+4%2F5%29%7D%5D+%3D+

Answer 47

math, you used 5 instead of 1/5

Answer 48

o.O

Answer 49

you made me feel bad again

Answer 50

but, will that matter? as I used 2 instead of 1/2

Answer 51

I think, I'm on a wrong track

Answer 52

Yeah, there's a bit of an error in your third step far, far above when removing the negative signs though.

Answer 53

So in the exponent of 2 we had \[-\log_3(-\log_5(x^2-4/5))\] but I don't think that those signs cancel each other out.

Answer 54

kainui method is correct

also i just realized,
x^2 -4/5

say x = 0
it'll be -4/5

which is not allowed

Answer 55

OH YES!

Answer 56

THanks @hartnn
@Kainui and @No.name for this question.