# Homework Help: Tricky trignometry

1. Jul 15, 2006

### vijay123

hi,
i got 2 problems that are really hard to solve. i dont 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:

2. Jul 15, 2006

### 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?

3. Jul 15, 2006

### Prince Stephen Ranji

What trignometry in these problems

4. Jul 15, 2006

### 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

5. Jul 15, 2006

### vijay123

yes, i have tried all of them, they dont seem to work!

6. Jul 15, 2006

### vijay123

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

7. Jul 15, 2006

### 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 dont know how to go on without using a calc. hope some1 has a better solution to the problem.

8. Jul 15, 2006

### nazzard

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

9. Jul 15, 2006

### 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 doesnt do any good.i come back to the same question.

10. Jul 15, 2006

### nazzard

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

Regards,

nazzard

11. Jul 15, 2006

### 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

12. Jul 15, 2006

### 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

13. Jul 15, 2006

### vijay123

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

14. Jul 15, 2006

### vijay123

the log 10 is 1 actually.....

15. Jul 15, 2006

### nazzard

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

Last edited: Jul 15, 2006
16. Jul 15, 2006

### 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:
ab + c = ab ac
ab - c = ab / ac
(ab)c = abc
(a / b)c = ac / bc
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 ax = 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. :)

Last edited: Jul 15, 2006
17. Jul 15, 2006

### nazzard

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

18. Jul 15, 2006

### vijay123

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

19. Jul 15, 2006

### vijay123

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

20. Jul 15, 2006

### vijay123

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