Solve Tricky Trigonometry Problems - Help Needed

[vijay123]

hi,
i got 2 problems that are really hard to solve. i don't know wether they are trick question or they just can be incorperated in a formula.i have taken days to solve them but in vain. please help!

a) 10(3^(2x+1)) = 2^(4x-3)

b) (2/3x)^(log2) = (9x)^(log3)

thanks you very much:tongue:

 

[Mindscrape]

Wait, this is just solve for x, right? Trigonometry?

I can help, but what have you done so far? Have you used all the properties of logs?

 

[Prince Stephen Ranji]

What trignometry in these problems

 

[nazzard]

Hello vijay,

this problem is fairly hard to solve with trigonometry

Have you tried the following identities?

$$\log(xy)=\log(x)+\log(y)$$

$$\log\left(\frac{x}{y}\right)=\log(x)-\log(y)$$

$$\log\left(x^y\right)=y\log(x)$$

Regards,

nazzard

 

[vijay123]

yes, i have tried all of them, they don't seem to work!

 

[vijay123]

yes, the rpoblem is trignometric, but all my efforts fail.

 

[vijay123]

and for info. this sum is a non calculator one...so...
anyway..for the first question, it might be helpful to expand out 10 into 5*2(accoring to me). the brought the 2 to the other side and the after further simplying, found out that (16/9)^x = 240...i don't know how to go on without using a calc. hope some1 has a better solution to the problem.

 

[nazzard]

yes, i have tried all of them, they don't seem to work!

Let's take a look at problem a)

$$10=\frac{2^{(4x-3)}}{3^{(2x+1)}}$$

$$\log10=\log\left(\frac{2^{(4x-3)}}{3^{(2x+1)}}\right)$$

Can you apply the second and third identity I posted earlier here?

Regards,

nazzard

 

[vijay123]

how would that help. the equation you have given is just the manipulated version of the problem. i tried to simplfy it, but it doesn't do any good.i come back to the same question.

 

[nazzard]

how would that help. the equation you have given is just the manipulated version of the problem. i tried to simplfy it, but it doesn't do any good.i come back to the same question.

Ok, so I have to ask what you want to achieve actually. Do you need an expression for x or something else?

Regards,

nazzard

 

[vijay123]

i would like to simplfy the rpbolem, making x the subject, were the whole equation is representated in exponential format, were x is equal to something in exponential form were there are no logarithms or any variable present.
regards
vijay

 

[nazzard]

I'll try to show the next steps I'd take:

$$\log10=\log(2^{(4x-3)})-\log({3^{(2x+1)}})$$

$$\log10=(4x-3)\log(2)-(2x+1)\log(3)$$

Can you solve it for x?

Regards,

nazzard

 

[vijay123]

yes yes..thats exactly wut i did...but now were near solution...lol...

 

[vijay123]

the log 10 is 1 actually...

 

[nazzard]

the log 10 is 1 actually...

Hang on...what makes you so sure that the base was chosen to be 10?

What do you mean with nowhere near the solution? Aren't these steps taking you closer and closer to it

You've tried this step yet?

$$\log10=4x\log(2)-3\log(2)-2x\log(3)-\log(3)$$

$$\log10=x\log(2^4)-\log(2^3)-x\log(3^2)-\log(3)$$

Regards,

nazzard

 

[VietDao29]

Ok, the simple way is to simplify the equation you get, then separate the unknown x into one side, and solve the equation from there.
I'll start off the first problem for you.
$$10 \times 3 ^ {2x + 1} = 2 ^ {4x - 3}$$
This looks too complicated, we must then use the following formulae to simplify it a bit:
a^{b + c} = a^b a^c
a^{b - c} = a^b / a^c
(a^b)^c = a^bc
(a / b)^c = a^c / b^c
We have:
$$\Leftrightarrow 10 \times 3 ^ {2x} \times 3 = \frac{2 ^ {4x}}{2 ^ 3}$$
$$\Leftrightarrow 30 \times {(3 ^ {2})} ^ x = \frac{{(2 ^ {4})} ^ x}{8}$$
$$\Leftrightarrow 30 \times 9 ^ x = \frac{16 ^ x}{8}$$
This looks much less complicated, right?
So, let's isolate x:
$$\Leftrightarrow \frac{9 ^ x}{16 ^ x} = \frac{1}{8 \times 30}$$

$$\Leftrightarrow \left( \frac{9}{16} \right) ^ x = \frac{1}{240}$$
Now if a^x = b, then how can you solve x in terms of a, and b?
Hint: Use logarithm. :)
The second one can be done exactly in the same way.
Can you go from here? :)
----------------
By the way, it's certainly not a trigonometry problem. :)

 

[nazzard]

Ok, the simple way is to simplify the equation you get, then separate the unknown x into one side, and solve the equation from there. SNIP

Very nice!

Vijay, if you'd like to compare VietDao's solution with mine, here is the result I get:

$$x=\frac{\log(10*3*2^3)}{\log\left(\frac{2^4}{3^2}\right)}$$

Both ways should give the same values for x.

Regards,

nazzard

 

[vijay123]

lol...dats wut i got in the first place!

 

[vijay123]

how do you simply dat?...dats the problem...i got the exact same ans.

 

[vijay123]

lol...sry...not trignometry...logarithms...

 

[vijay123]

hey guys...do you think answer is the best solution...cause i have always been sceptical of such logarithm answers...do you think that this is the best possible solution

 

[Mindscrape]

There is no such thing as a best possible solution. You have two ways of attaining a solution, and you can pick either one.

 

[vijay123]

so..this naswer is fine?..

 

[vijay123]

anyway...lets discuss the second question..i got REAL problems with that one.

 

[VietDao29]

anyway...lets discuss the second question..i got REAL problems with that one.

?
Am I that easy to be fooled? What do you mean by discuss?
I used the first problem as an example, so that you (yes, it's you) can follow the similar steps to arrive to the correct answer. Of course, I did not intend to give you the COMPLETE solution.
AFAIK, you have not shown any of your work. All you said was something like: 'that's exactly wut I did', or even worse 'I got the same answer'. Let's face the truth then, if you did get the correct answer, why bother to post it here?
*completely disappointed, and frustrated* :grumpy: :grumpy: :grumpy:
If you don't mind, you can read some of our forum rules https://www.physicsforums.com/showthread.php?t=28.

 

[vijay123]

lol...i got the answer...at last...thanks for your help guys!

 

[vijay123]

fine...you make me angry...first of all...i can't show my work because i don't know hwo to you the log thingee...and if you want to see my work..then teach me how to..lets see how disappointed you get then!

 

[vijay123]

you teach me...and i ll prove to show you my working, for both problmes...get it?...anyway...i couldn't show you because the steps get real complex...especially wen you type it all in the computer...so please teach me the rpgram you did and i ll show you...i am not as dumb as wut u think!

 

[HallsofIvy]

Apparently you don't understand the word "teach". You want someone to do the problem for you and then you will "show you my working" by repeating it back?
Both nazzard and VietDao29 did your first problem for you- showing you HOW to do it, step by step, and your response was "I got the exact same answer but I don't know how to do it". Since you got the "exact same answer", how about showing your work on that problem and we can see what YOU did.

The second problem is exactly the same as the first. Try it!

 

[vijay123]

fine!i ll show you. here how it goes...
1)10(3^(2x+1)=2^(4x-3)
5(3^(2x+1)=2^(4x-4)
5*3*2^4=2^4x/3^2x
240=2^4x/3^2x
hence x=0.5((log240))/(log(4/3))

happy...hope i am rite...
anyway...for the second answer...i got mine as x=2/9...
check if mine is correct..

 

[vijay123]

anyway...i didnt mean to start any fight...it maybe some people mis read wut i wrote which caused the whole rpblem...i am truly saying that i got the same answer, just that i thought it wus wrong...since the answer is in a logarithmic form!hence, i thought that it would be impossible to simplify the problem...hence i sought the help of you guys...

 

[vijay123]

if you guys still don't believe me...then go check post no. 7! 7...yes...number SEVEN of this thread. i got the answer bfore, jus that i thought it wus wrong...any more arguments to continue?

 

[VietDao29]

...
240=2^4x/3^2x
hence x=0.5((log240))/(log(4/3))

This is fine, but you can simplify it a little bit further. Using the Change base formula:
$$\log_{a} b = \frac{\log_c b}{\log_c a}$$, we have:
$$x = \frac{1}{2} \times \frac{\log 240}{\log \left( \frac{4}{3} \right)} = \frac{1}{2} \times \log_{\frac{4}{3}} 240$$.
------------------
Or even faster, given that a^x = y, the unknown x is just x = log_ay
So you can do like this:
$$240 = \frac{2 ^ {4x}}{3 ^ {2x}}$$
$$\Leftrightarrow 240 = \frac{16 ^ {x}}{9 ^ {x}}$$
$$\Leftrightarrow 240 = \left( \frac{16}{9} \right) ^ x$$
$$\Leftrightarrow x = \log_{\frac{16}{9}} 240$$
----------------
Is the log in your second problem is in base e or base 10?
And by the way, there's no fight here. You didn't start any. :)

 

[vijay123]

uhh...tis base 10...
thanks for the simplification...

 

[VietDao29]

Ok, let's move on to the second question then:

anyway...for the second answer...i got mine as x=2/9...

How did you get x = 2 / 9? It does not look quite right to me.
It can be done exactly the same as the first problem. So would you mind if you show us how you get x = 2 / 9? Or do you need some help with this one also?
Just give it a try, and see what you get. If you get stuck somewhere, just podt it here.