Two Exponential Equations, and Two Triq Questions!

**theqwertykid** · December 12th 2008, 09:53 AM

1. $9^x+4*3^x - 3 = 0$

2. $5^x = 4^x+5$ <-----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)

**o_O** · December 12th 2008, 11:02 AM

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#1:

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

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

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

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

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

Now these will come in handy:

$\cos (a + b) = \cos a \cos b - \sin a \sin b$
$\cos (2a) = 2\cos^2 a - 1$
$\sin (2a) = 2\sin a \cos a$
$\sin^2 (a) = 1 - \cos^2 a$

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

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

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

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