Solve Exponential Function: 2 Questions | Help Needed

[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

 

[quantumdude]

Could you please show us what you've tried?

 

[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
*the answer was 5/4*

 

[love_joyously]

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

 

[quantumdude]

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.

 

[Jameson]

I'm trying to follow your work, but it's very difficult to read.

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

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

 

[love_joyously]

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

 

[love_joyously]

i tried it but i keep eliminating my variable

 

[quantumdude]

i tried it but i keep eliminating my variable

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

 

[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

?

 

[quantumdude]

Therefore, 4x-3/8x-3 = 1/2

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.

 

[love_joyously]

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

wat now?

 

[quantumdude]

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.

 

[djeipa]

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

wat now?

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

 

[love_joyously]

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

 

[love_joyously]

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

 

[djeipa]

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

 

[quantumdude]

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

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

 

[Ouabache]

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

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 I get, x = -1, 2