# Exponential Function

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

Staff Emeritus
Gold Member
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

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

Staff Emeritus
Gold Member
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.

Gold Member
MHB

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

Staff Emeritus
Gold Member
love_joyously said:
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

?

Staff Emeritus
Gold Member
love_joyously said:
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?

Staff Emeritus
Gold Member
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
love_joyously said:
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..

Staff Emeritus
Gold Member
djeipa said:
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.

Homework Helper
love_joyously said:
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 [Broken] I get, x = -1, 2

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