Calculus - Exponential functions

[Klthompson]

I am having a hard time solving this equation for x :

3^2x - 12*3^x + 27 = 0

I've gone this far:

3^2x -3^1*4*3^x + 3^3 = 0

I can't figure out what to do to the four to get it to a base of 3...[/list][/code]

 

[pka]

3<SUP>2x</SUP> - 12*3<SUP>x</SUP> + 27 = 0
If z=3<SUP>x</SUP> we have z<SUP>2</SUP> - 12z + 27 = 0 to solve.
Then use logs.

 

[Klthompson]

pka---that really confused me. what is "z" supposed to be? I'm solving for "x"

 

[pka]

(z−9)(z−3)=0 gives you z=9 & z=3.
If z=9 then 3<SUP>x</SUP>=9.
NOW SOLVE FOR x.

 

[Klthompson]

okay...

so (z-9)(z-3) = 0

z = 3, 9

so, 3^2 = 3^2x and x = 1

12*3 = 12*3^x and x = 1

so for the problem, x is equal to 1

AND

9^2 or 81 = 3^2x so 3^2 * 3^x = 81 and x = 2

so the answer is then 1, or 2....is that right?

 

[stapel]

...the answer is then 1, or 2....is that right?

To check the answer to any "solving" exercise, plug the solution back into the original problem, and see if it checks. So plug "1" in for "x" and see if the equation works out. Do the same with "2".

Eliz.

 

[soroban]

Hello, Klthompson!

\(\displaystyle 3^{2x}\,-\,12\cdot3^x\,+\,27\:=\:0\)

pka is correct . . . except we don't need logs.

We have: .\(\displaystyle (3^x)^2\,-\,12(3^x)\,+\,27\:=\:0\) . . . a quadratic in \(\displaystyle 3^x.\)

Let \(\displaystyle z\,=\,3^x\)

Then we have: .\(\displaystyle z^2\,-\,12z\,+\,27\:=\:0\)

. . which factors: .\(\displaystyle (z \,-\,3)(z\,-\,9)\:=\:0\)

. . and has roots: .\(\displaystyle z\,=3,\,9\)


Since \(\displaystyle z\,=\,3^x\), we have these two equations to solve:

. . \(\displaystyle 3^x\,=\,3\,=\,3^1\;\;\Rightarrow\;\;x\,=\,1\)

. . \(\displaystyle 3^x\,=\,9\,=\,3^2\;\;\Rightarrow\;\;x\,=\,2\)

 

[Klthompson]

Staple and Soroban----Thank you for the reply. As you can see by my work, I had already "plugged" in 1 and 2 for x and came up with the correct answers before you replied. Thanks for all of the help.