1. ## [SOLVED] Exponential Functions

I need help with 2 problems, please. I'm not sure how to solve them.

#1:
$\displaystyle 4x^{2}(2^{x}) - 9(2^{x}) = 0$

#2:
$\displaystyle \frac{(x+4)10^{x}}{x-3} = 2x(10^{x})$

Thanks for the help! I haven't done this kind of math in awhile, and I'm just confused.

2. Originally Posted by juicysharpie
I need help with 2 problems, please. I'm not sure how to solve them.

#1:
$\displaystyle 4x^{2}(2^{x}) - 9(2^{x}) = 0$

#2:
$\displaystyle \frac{(x+4)10^{x}}{x-3} = 2x(10^{x})$

Thanks for the help! I haven't done this kind of math in awhile, and I'm just confused.
#1 Factorise: $\displaystyle 2^x (4x^2 - 9) = 0$. Now try solving for x.

#2 Re-arrange: $\displaystyle \frac{(x+4)10^{x}}{x-3} - 2x(10^{x}) = 0$.

Now factorise: $\displaystyle 10^x \left( \frac{x + 4}{x - 3} - 2x \right) = 0$

$\displaystyle \Rightarrow 10^x \left( \frac{x + 4}{x - 3} - \frac{2x (x - 3)}{x - 3} \right) = 0$

$\displaystyle \Rightarrow 10^x \left( \frac{x + 4 - 2x(x - 3)}{x - 3}\right) = 0$

$\displaystyle \Rightarrow 10^x \left( \frac{-2x^2 + 7x + 4}{x - 3}\right) = 0$

$\displaystyle \Rightarrow 10^x \left( \frac{(2x + 4)(4 - x)}{x - 3}\right) = 0$

Now try solving for x.

3. Thanks for the help!

I just wanted to check my answers if that's O.K.

#1: $\displaystyle x = \frac{3}{2}$

and

#2 $\displaystyle x = -2$ or $\displaystyle -4$

Are those correct?

4. Originally Posted by juicysharpie
Thanks for the help!

I just wanted to check my answers if that's O.K.

#1: $\displaystyle x = \frac{3}{2}$

and

#2 $\displaystyle x = -2$ or $\displaystyle -4$

Are those correct?
Not quite.

#1: $\displaystyle 4x^2 - 9 = 0 \Rightarrow (2x - 3)(2x + 3) = 0 \Rightarrow x = \pm \frac{3}{2}$. Note: $\displaystyle 2^x = 0$ has no real solution.

#2: $\displaystyle 2x + 4 = 0 \Rightarrow x = -2$ or $\displaystyle 4 - x = 0 \Rightarrow x = 4$ (NOT $\displaystyle -4$).