Question

Some questions on Logs for people to try

Answer 1

Evaluate:
\[\log_{\tan(20)}\tan(70) \]
\[\log_{2-\sqrt{3}}(2+\sqrt{3})\]
\[6+\log_{\frac{3}{2}}(\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}...}})\]
Find the solution of the following equation:
\[7^{\log_{7}x}+2x+9=0\]
Evaluate:
\[\log_{10}\tan(5) \times \log_{10}\tan(9) \times \log_{10}\tan(13)...\log_{10}\tan(61)\]

All measures in the tangent functions are in degrees

Answer 2

@ganeshie8

Answer 3

@amistre64

Answer 4

@dan815

Answer 5

this is something you need help with ?

Answer 6

Nope

Answer 7

ok so it is just fun fun!

Answer 8

I found these questions, thought I'd share

Answer 9

Give them a try

Answer 10

For the 2nd one we can make use of the fact that:\[(2+\sqrt{3})(2-\sqrt{3})=-1\]

Answer 11

For the 4th one we can make use of the fact that:\[7^{\log_7x}=x\]

Answer 12

For the first one we can do:\[\tan(70)=\tan(90-20)=\frac{\tan(90)-\tan(20)}{1+\tan(90)\tan(20)}\]\[=\frac{1-\frac{\tan(20)}{\tan(90)}}{\frac{1}{\tan(90)}+\tan(20)}=\frac{1-0}{0+\tan(20)}=\frac{1}{\tan(20)}\]

Answer 13

The 3rd and 5th ones look more involved

Answer 14

omg @asnaseer I have been doing some crazy things for the 1st one

Answer 15

you made it so easy

Answer 16

Interesting, but even the 4th is tricky :)

Answer 17

You could have also used \[\tan(90-\theta)=\cot(\theta)=\frac{1}{\tan(\theta)}\]

Answer 18

@Nishant_Garg - why is the 4th one tricky? Can't you just use the fact that \(7^{\log_7x}=x\)?

Answer 19

Yes you can but even then there's a trick, I hope you can figure it out :)

Answer 20

Hang on, I just had a thought on the 3rd one, we can say:\[x=\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}...}}\]The if we square both sides we get:\[x^2=\frac{1}{18}(4-x)\]\[\therefore18x^2+x-4=0\]\[\therefore(2x+1)(9x-4)=0\]\[\therefore x=\frac{4}{9}\]since it cannot be negative. The solution then becomes simple

Answer 21

Nice, nice!

Answer 22

On the fourth one I get:\[7^{\log_{7}x}+2x+9=0\]\[\therefore x+2x+9=0\]Which leads to \(x=-3\)

Or have I missed something???

Answer 23

unless we are saying it cannot be negative

Answer 24

You didn't miss anything but that is not the correct answer :P

Answer 25

because of the log?

Answer 26

Yep!

Answer 27

Sneeky! :)

Answer 28

So what is the solution?hm?

Answer 29

So it becomes something like:\[|x|+2x+9=0\]

Answer 30

I would think there is no real solution

Answer 31

should we consider complex solutions?

Answer 32

or maybe:\[\sqrt{x^2}+2x+9=0\]

Answer 33

but log(x) has domain for x>0

Answer 34

Indeed, there is no solution, I didn't think about complex, however

Answer 35

well when we talk about real numbers

Answer 36

that leads to:\[x=3(1\pm i\sqrt{2})\]

Answer 37

I think complex would be super complex :p

Answer 38

@asnaseer is that the complex solution to the 4th one?

Answer 39

yes - I think so

Answer 40

how did you get that?

Answer 41

computer calculator thingy?

Answer 42

paper and pencil :)

Answer 43

but I am not sure I did the steps correctly

Answer 44

in fact I am sure I have made a mistake on that one :(

Answer 45

Try the 5th :)

Answer 46

Oh yes - still one left :)

Let me try...

Answer 47

@jigglypuff314 hey try these questions :p

Answer 48

I am currently thinking along the lines of using something like:\[10^{\log_{10}\tan(x)}=\tan(x)\]

Answer 49

Probably easy for you guys, but try this problem:

Evaluate \[\Huge\ddots\log_{\log_{\log_{\log_e4}4}4}4^{.^{.^.}}\]

Answer 50

Is it 2?

Answer 51

None :)

Answer 52

Is it extending on both the sides?

Answer 53

Yeah

Answer 54

Can you explain how you got 2 though?

Answer 55

@geerky42 - your equation has just melted my brain! :)

Answer 56

My guess is that we need to start with the inner most \(\log_e4\)

Answer 57

I did something like this:
\[x=\log_{x}4\]\[x^x=4\]\[x=2\]
But then there's the base e in the middle that makes no sense to me :)

Answer 58

hey I cheated http://www.wolframalpha.com/input/?i=product%28log_%2810%29%285%2B4i%29%2Ci%3D0..14%29
wolfram doesn't seem to make exact form cute at all for that one problem

Answer 59

To "wording" this problem better.
Let \(a_n = log_{a_{n-1}}4\), where \(a_1 = \log_{e}4\), evaluate \(\lim_{n\to\infty} a_n\)

Answer 60

HINT: Check for existence of this limit.

Answer 61

Anyone?

Answer 62

I'm thinking about it

Answer 63

I don't know, I'm lost O.o

Answer 64

Your question is definitely harder

Answer 65

@asnaseer @freckles How are you doing with this problem?

Answer 66

@geerky42 - too difficult for me but I am very interested in learning about how to solve this

Answer 67

\[v^j=4 \\ u^v=4 \\ x^u=4 \\ e^x=4 \]
I have these equations where j is the whole expression
like I started with the bottom thingy called it x
like I did this:
\[\text{ Let } x=\log_e(4) \\ \text{ then } u=\log_x(4) \\ \text{ then } v=\log_u(4) \\ \text{ then } j=\log_v(4)\]

Answer 68

but still playing with it

Answer 69

okay, want me to post answer anyway? @freckles

Answer 70

like I can solve for x
then u then v then j (which is the main objective)

am I going about it the right way?

Answer 71

or a way that will work I mean

Answer 72

if not yes post the answer

Answer 73

I would say you are in correct path. @freckles

Answer 74

@freckles that reminds of those dolls where you put the smaller doll inside the bigger one and even smaller one inside that small one

Answer 75

:)

Answer 76

Yep, that one!

Answer 77

Russian Dolls :)

Answer 78

\[\text{ Let } x=\log_e(4) \\ \text{ then } u=\log_x(4) \\ \text{ then } v=\log_u(4) \\ \text{ then } j=\log_v(4) \\\ \text{ so we have } \]
\[v^j=4 \\ u^v=4 \\ x^u=4 \\ e^x=4\]
\[x=\ln(4) \\ (\ln(4))^u=4 \\ u \ln(\ln(4))=\ln(4) \\ u=\frac{\ln(4)}{\ln(\ln(4))} \\ (\frac{\ln(4)}{\ln(\ln(4))})^v=4 \\ v \ln(\frac{\ln(4)}{\ln(\ln(4))})=\ln(4) \\ v=\ln(4) \frac{1}{\ln(\frac{\ln(4)}{\ln(\ln(4))})}\]
and then lastly we solve for j
and just looks really ugly to me :(

Answer 79

You read my reworded problem, right? @freckles

Answer 80

\[v^j=4 \\ (\ln(4) \frac{1}{\ln(\frac{\ln(4)}{\ln(\ln(4))})})^j=4\]
no :p
I was too busy playing with the thingy I thought was a 2 sided exponent which didn't make sense to me at first
until I seen it as what a base inside a base inside a base

Answer 81

Ahh yeah as I thought.

I was kinda confused at what you are trying to do, lol.

Answer 82

I want to give up though
I'm ready to receive your answer with open arms

Answer 83

Okay. Answer is it doesn't exist!

If you decide to do brute force to see where value is approaching to as n grow in \(a_n\);

\[a_1 = 1.3863...\\a_2=4.2442...\\a_3=0.9590...\\a_4=-33.1217\]
Since \(a_4\) is negative, \(a_5\) doesn't exist, hence \(\displaystyle \lim_{n\to\infty}a_n\) doesn't exist.

Answer 84

D'oh! - I was hoping to be enlightened by some fancy method :)

Answer 85

Yeah I had tried manually solving it and I noticed the value first goes up and then goes down but I stopped afterwards because I couldn't find a base 1.3863 log calculator, didn't think it would go in negative....lol

Answer 86

haha, yeah solution is kinda let-down. but still pretty entertain to me.

Answer 87

@geerky42 can u solve the 5th question I originally posted?:P

Answer 88

@Nishant_Garg is the answer to the 5th one zero?

Answer 89

I've to go to bed in a few mins so I'd really like if that question is solved before I have to go, all the questions would be done then, but I'll keep the question open till tomorrow so others can have a look aswell

Answer 90

because one of terms is \(\log_{10}\tan(45)=0\)

Answer 91

Hahaha, yeah you did it.

Answer 92

phew! :D

Answer 93

omg wolfram is stupid :(
asanseer you outsmarted wolfram http://www.wolframalpha.com/input/?i=product%28log_%2810%29%285%2B4i%29%2Ci%3D0..14%29

Answer 94

Nice catch @asnaseer !

Answer 95

oh I forgot to put tangent :

Answer 96

I guess the complicated question posted by @geerky42 threw my brain into the correct gear to solve it! :D