# Exponential Function

1. Sep 22, 2005

### love_joyously

i'm having problems with two questions. Please help me! Thanks! i've tried everything but i can't solve them...

1) Solve: (3/4)^3x-2 * (4/3)^1-x = 9/16

2) Solve for x : 3(3^x) + 9(3^-x)=28

2. Sep 22, 2005

### Tom Mattson

Staff Emeritus
Could you please show us what you've tried?

3. Sep 22, 2005

### love_joyously

Well for no. 1 this is what i've tried but i keep getting the wrong answer:
*2nd step* 3^3x-2/ (2^2)^3x-2 * (2^2)^1-x/3^1-x = 9/16
3^3x-2/2^6x-4 * 2^2-2x^3^1-x = 9/16
2^-8x+6/3^-4x+3 = 3^2/2^4
Therefore, -8x+6/-4x+3 = 1/2
-16x +12 = -4x +3
-12x = -9
x = 3/4

4. Sep 22, 2005

### love_joyously

and for no.2 i have no idea what my next step is...

5. Sep 22, 2005

### Tom Mattson

Staff Emeritus
There's a much easier way.

First, recognize that $\frac{4}{3}=(\frac{3}{4})^{-1}$ and that $\frac{9}{16}=(\frac{3}{4})^2$.

Once you do that, all the bases will be the same. Then you can apply the rule $a^xa^y=a^{x+y}$ to simplify the left side, and then solve the equation.

You'll want to do something similar to reduce the left side of #2 to a single term.

6. Sep 22, 2005

### Jameson

Here is your second step in Latex form. Maybe you could rewrite the next steps.

$$\frac{3^{3x-2}}{(2^2)^{3x-2}}*\frac{(2^2)^{1-x}}{3^{1-x}}$$

7. Sep 22, 2005

### love_joyously

yeah, sure, ill try. Then ill let you know

8. Sep 22, 2005

### love_joyously

i tried it but i keep eliminating my variable

9. Sep 22, 2005

### Tom Mattson

Staff Emeritus
How? When you add the exponents, the variable does not cancel out.

10. Sep 22, 2005

### love_joyously

ok.. well this is what i did:

3^3x-2/2^6x-4 * 3^x-1/2^2x-2 = 3^2/2^4
3^4x-3/2^8x-6=3^2/2^4
Therefore, 4x-3/8x-3 = 1/2
*cross-multiply* 8x-6 = 8x-6
0=0

?????

11. Sep 22, 2005

### Tom Mattson

Staff Emeritus
This step is wrong.

You'll have better luck if you don't use different exponents for the numerator and denominator.

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

Now you can simply add the exponents on the left side, and solve for x.

12. Sep 22, 2005

### love_joyously

ook.. i get : (3/4)^4x-3 = (3/4)^2

wat now?

13. Sep 22, 2005

### Tom Mattson

Staff Emeritus
See how the bases are equal? That means that the exponents must be equal in order for the equation to hold. Set them equal, then solve for x.

14. Sep 22, 2005

### djeipa

I know know this problem,
4x-3=2
then x=5/4

15. Sep 22, 2005

### love_joyously

oh..ok!!! now i get it.. thanks!

16. Sep 22, 2005

### love_joyously

so for no. 2, what do u suggest me do?

17. Sep 22, 2005

### djeipa

let t=3^x then Solve for t : 3t^2 -28t + 9=0
Yeah..

18. Sep 22, 2005

### Tom Mattson

Staff Emeritus
That's what I would do, too. Solve that quadratic equation, then solve for x.

19. Sep 23, 2005

### Ouabache

I was trying to follow what you did to arrive at the quadratic expression.
It seems you multiplied the original equation by $3^x$

$3^x [ {3(3^x)+9(3^{-x})=28}]$

$3(3^x)^2 +9 = 28(3^x)$

$3(3^x)^2 -28(3^x) +9 = 0$

And If $t = 3^x$ then $3t^2-28t+9=0$

With a bit of http://www.deephousepage.com/smilies/OLA.gif [Broken] I get, x = -1, 2

Last edited by a moderator: May 2, 2017