# Still trigonometry

• Dec 12th 2012, 04:23 AM
dirtyharry
Still trigonometry
22x+1 - 5 * 2x - 12 = 0

If 2x is y, then it would be like this:

y2 - 5y - 12 = 0

Seems simple enough, but:

$\displaystyle \frac{-5 \pm \sqrt{5^{2} - 4 * 1 * -12 }}{2}$

equals:

$\displaystyle \frac{-5 \pm \sqrt{73 }}{2}$

And there is no string square root of 73. Is there a way to get around this, some trick perhaps?

Next one is:

$\displaystyle 5^{x^{2} - 3x + 1} = \frac{1}{5}$

I believe I have to solve the $\displaystyle x^{2} - 3x + 1$ part first, and so I tried, but encountered the same problem as above: The square root ends up like a decimal - which shouldn't happen :/

And neither last nor least, is:

$\displaystyle (\frac{1}{2})^{x} * (\frac{2}{3})^{x+1} * (\frac{3}{4})^{x+2} = 6$

I don't even know how to start this one. I believe that's about it from exponentials, but I have to cover two more topics about trigonometry :O

Thanks
• Dec 12th 2012, 04:40 AM
MarkFL
Re: Still trigonometry
You have made an error in your substitution in the first problem. Your first term should be $\displaystyle 2y^2$. Do you see why?
• Dec 12th 2012, 05:11 AM
dirtyharry
Re: Still trigonometry
I do now, and it makes sense. idk why I didn't see it before.

• Dec 12th 2012, 05:31 AM
MarkFL
Re: Still trigonometry
For the second one, rewrite the right side as $\displaystyle 5^{-1}$, then equate exponents and solve for x.

For the third, I would rewrite the equation as:

$\displaystyle \frac{2}{3}\cdot\left(\frac{3}{4} \right)^2\cdot\left(\frac{1}{2} \right)^x\left(\frac{2}{3} \right)^x\left(\frac{3}{4} \right)^x=6$

$\displaystyle \left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4} \right)^x=6\cdot\frac{8}{3}$

Can you proceed from here?
• Dec 12th 2012, 06:12 AM
dirtyharry
Re: Still trigonometry
Quote:

Originally Posted by MarkFL2
For the second one, rewrite the right side as $\displaystyle 5^{-1}$, then equate exponents and solve for x.

For the third, I would rewrite the equation as:

$\displaystyle \frac{2}{3}\cdot\left(\frac{3}{4} \right)^2\cdot\left(\frac{1}{2} \right)^x\left(\frac{2}{3} \right)^x\left(\frac{3}{4} \right)^x=6$

$\displaystyle \left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4} \right)^x=6\cdot\frac{8}{3}$

Can you proceed from here?

I believe I can. I'm not familiar with how you got the 8/3. I know it's from $\displaystyle \frac{2}{3}\cdot\left(\frac{3}{4} \right)^2$ but I can't see how you got there.
If the right answer is -2, then I got it right.
• Dec 12th 2012, 06:22 AM
MarkFL
Re: Still trigonometry
Quote:

Originally Posted by dirtyharry
I believe I can. I'm not familiar with how you got the 8/3. I know it's from $\displaystyle \frac{2}{3}\cdot\left(\frac{3}{4} \right)^2$ but I can't see how you got there.
If the right answer is -2, then I got it right.

$\displaystyle \frac{2}{3}\cdot\left(\frac{3}{4} \right)^2=\frac{2}{3}\cdot\frac{9}{16}=\frac{3}{8}$

So, then I multiplied both sides by $\displaystyle \frac{8}{3}$. And yes, $\displaystyle x=-2$ is the correct answer, since we find:

$\displaystyle \left(\frac{1}{4} \right)^x=16=\left(\frac{1}{4} \right)^{-2}$
• Dec 12th 2012, 06:38 AM
dirtyharry
Re: Still trigonometry
Thanks. I'm doing logorythms now. Can you help me with this: What's X if:
X = $\displaystyle X = 5^{2 + 2log_{5}4}$
• Dec 12th 2012, 06:43 AM
MarkFL
Re: Still trigonometry
I would write:

$\displaystyle X=5^{2+2\log_5(4)}=5^2\cdot5^{\log_5(4^2)}=?$