Still trigonometry

**dirtyharry** · December 12th 2012, 04:23 AM

2^2x+1 - 5 * 2^x - 12 = 0

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

y² - 5y - 12 = 0

Seems simple enough, but:

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

equals:

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

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

Next one is:

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

I believe I have to solve the $x^{2} - 3x + 1$ 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:

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

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

**MarkFL** · December 12th 2012, 04:40 AM

You have made an error in your substitution in the first problem. Your first term should be $2y^2$ . Do you see why?

**dirtyharry** · December 12th 2012, 05:11 AM

I do now, and it makes sense. idk why I didn't see it before.

What about the other two?

**MarkFL** · December 12th 2012, 05:31 AM

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

For the third, I would rewrite the equation as:

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

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

Can you proceed from here?

**dirtyharry** · December 12th 2012, 06:12 AM

Originally Posted by MarkFL2

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

For the third, I would rewrite the equation as:

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

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

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 $\frac{2}{3}\cdot\left(\frac{3}{4} \right)^2$ but I can't see how you got there.
If the right answer is -2, then I got it right.

**MarkFL** · December 12th 2012, 06:22 AM

Originally Posted by dirtyharry

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

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

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

$\left(\frac{1}{4} \right)^x=16=\left(\frac{1}{4} \right)^{-2}$

**dirtyharry** · December 12th 2012, 06:38 AM

Thanks. I'm doing logorythms now. Can you help me with this: What's X if:
X = $X = 5^{2 + 2log_{5}4}$

**MarkFL** · December 12th 2012, 06:43 AM

I would write:

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

Thread: Still trigonometry

LinkBack

Thread Tools

Display

Still trigonometry

Re: Still trigonometry

Re: Still trigonometry

Re: Still trigonometry

Re: Still trigonometry

Re: Still trigonometry

Re: Still trigonometry

Re: Still trigonometry

Similar Math Help Forum Discussions

Trigonometry to Memorize, and Trigonometry to Derive

Trigonometry help plz!!!

trigonometry

Trigonometry

Trigonometry

Search Tags