- #1

love_joyously

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

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

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- Thread starter love_joyously
- Start date

- #1

love_joyously

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

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

- #2

quantumdude

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Could you please show us what you've tried?

- #3

love_joyously

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*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*

- #4

love_joyously

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

- #5

quantumdude

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

Once you do that, all the bases will be the same. Then you can apply the rule [itex]a^xa^y=a^{x+y}[/itex] 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

Jameson

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

[tex]\frac{3^{3x-2}}{(2^2)^{3x-2}}*\frac{(2^2)^{1-x}}{3^{1-x}}[/tex]

- #7

love_joyously

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

- #8

love_joyously

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i tried it but i keep eliminating my variable

- #9

quantumdude

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love_joyously said:i tried it but i keep eliminating my variable

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

- #10

love_joyously

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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

quantumdude

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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.

[tex]\left(\frac{3}{4}\right)^{3x-2}\left(\frac{3}{4}\right)^{x-1}=\left(\frac{3}{4}\right)^2[/tex]

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

- #12

love_joyously

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

wat now?

wat now?

- #13

quantumdude

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- #14

djeipa

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I know know this problem,love_joyously said:ook.. i get : (3/4)^4x-3 = (3/4)^2

wat now?

4x-3=2

then x=5/4

- #15

love_joyously

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

- #16

love_joyously

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

- #17

djeipa

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

Yeah..

Yeah..

- #18

quantumdude

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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.

- #19

Ouabache

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I was trying to follow what you did to arrive at the quadratic expression.love_joyously said:2) Solve for x : 3(3^x) + 9(3^-x)=28

It seems you multiplied the original equation by [itex]3^x[/itex]

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

[itex] 3(3^x)^2 +9 = 28(3^x) [/itex]

[itex] 3(3^x)^2 -28(3^x) +9 = 0 [/itex]

And If [itex] t = 3^x [/itex] then [itex] 3t^2-28t+9=0 [/itex]

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

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