# Two Exponential Equations, and Two Triq Questions!

• Dec 12th 2008, 09:53 AM
theqwertykid
Two Exponential Equations, and Two Triq Questions!
1.$\displaystyle 9^x+4*3^x - 3 = 0$

2. $\displaystyle 5^x = 4^x+5$ <-----its 4^x+5, 4 is raised to the x+5!!! Can't figure out how to make it look that way.. =/

3. A formula for cos(3X) in terms of cosX alone.

4. The period of 3cos(5pix+7)
• Dec 12th 2008, 11:02 AM
o_O
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#1:

\displaystyle \begin{aligned}9^x + 4 \cdot 3^x - 3 & = 0 \\ \left(3^2\right)^x + 4 \cdot 3^x - 3 & = 0 \\ \left(3^x\right)^2 + 4 \cdot 3^x - 3 & = 0 \end{aligned}

If it helps, imagine $\displaystyle y = 3^x$. Then we get an easier problem: $\displaystyle y^2 + 4y - 3 = 0$

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#2: Start by taking the natural logarithm of both sides ..

\displaystyle \begin{aligned} {\color{red}\ln \ }5^x & = {\color{red}\ln \ }4^{x+5} \\ x \ln 5 & = (x+5) \ln 4 \qquad \text{ Since: } \ln a^b = b\ln a\\ & \ \ \vdots \end{aligned}

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#3: $\displaystyle \cos (3x) = \cos (x + 2x)$

Now these will come in handy:
• $\displaystyle \cos (a + b) = \cos a \cos b - \sin a \sin b$
• $\displaystyle \cos (2a) = 2\cos^2 a - 1$
• $\displaystyle \sin (2a) = 2\sin a \cos a$
• $\displaystyle \sin^2 (a) = 1 - \cos^2 a$

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#4: $\displaystyle 3\cos \left(5\pi x + 7\right) \ = \ 3\cos \left( 5\pi \left(x + \frac{7}{5\pi}\right)\right)$

If you remember, if you have a sinusoidal wave in the form: $\displaystyle y = A\cos (B (x + C)) + D$

The period is given by: $\displaystyle \frac{2\pi}{B}$

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