solve 3^(2x^2+1) - 3^(x^2+2) = 3^(x^2) - 3 and

[oded244]

 

[galactus]

Re: help with exponentials

For the first one, since the bases are the same, concentrate on the exponents

$\displaystyle 2x^{2}+1-(x^{2}+2)=x^{2}-1$

What do you get?.

 

[oded244]

Re: help with exponentials

hmm.. i see.
why can't i put a t=9^x ? im getting +-0.5 when i do that.

 

[stapel]

why can't i put a t=9^x ? im getting +-0.5 when i do that.

Um... what...? What do you mean by "putting a t = 9[sup:30pyum36]x[/sup:30pyum36]"? Putting it where? Who said you can't? (I don't see any previous reference to this.) For what are you "getting 0.5"?

Please reply with clarification, showing all of your work and reasoning. Thank you!

Eliz.

 

[oded244]

Re: help with exponentials

that's what i did for the first one:

3^x^2=t
t^2*3-t*3^2=t-3
3t^2-10t+3=0
t=3 t=1/3

3^x^2=3 .....= 1/2

 

[wjm11]

Re: help with exponentials

3^x^2=3 .....= 1/2

Are you saying that x = 1/2 is a solution? Have you tried plugging that back in to the original equation to check it?

If 3^(x^2) = 3^1, then x^2 = 1. Therefore,
x = +/- 1

 

[soroban]

Re: help with exponentials

Hello, oded244!

$\displaystyle 3^{2x^2+1} - 3^{x^2+2} \;=\;3^{x^2} - 3$

$\displaystyle \text{We have: }\;3^{2x^2+1} - 3^{x^2+2} - 3^{x^2} + 3 \;=\;0$

$\displaystyle \text{Factor: }\;3^{x^2+2}\left(3^{x^2-1} - 1\right) - 3\left(3^{x^2-1} - 1\right) \;=\;0$

$\displaystyle \text{Factor: }\;\left(3^{x^2+2} - 3\right)\,\left(3^{x^2-1} - 1\right) \;=\;0$


$\displaystyle \text{We have: }\;3^{x^2+2} - 3 \:=\:0 \quad\Rightarrow\quad 3^{x^2+2}\:=\:3^1$

. . \(\displaystyle x^2+2 \:=\:1 \quad\Rightarrow\quad x^2\:=\:-1\quad\hdots\text{ no real roots}\)


$\displaystyle \text{And we have: }\;3^{x^2-1}-1 \:=\:0 \quad\Rightarrow\quad 3^{x^2-1} \:=\:1 \quad\Rightarrow\quad 3^{x^2-1} \:=\:3^0$

. . $\displaystyle x^2-1 \:=\:0\quad\Rightarrow\quad x^2 \:=\:1 \quad\Rightarrow\quad\boxed{ x \:=\:\pm 1}$

 

[soroban]

Re: help with exponentials

Hello again, oded244!

The second problem is a killer . . .

$\displaystyle 4^x - 3^{\frac{2x-1}{2}} \;=\;3^{\frac{2x+1}{2}} - 2^{2x-1}$

$\displaystyle \text{We have: }\;\left(2^2\right)^x + 2^{2x-1} \;=\;3^{\frac{2x+1}{2}} + 3^{\frac{2x-1}{2}} \quad\Rightarrow\quad 2^{2x} + 2^{2x-1} \;=\;3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}$

$\displaystyle \text{Factor: }\;2^{2x-1}(2 + 1) \;=\;3^{x-\frac{1}{2}}(3 + 1) \quad\Rightarrow\quad 2^{2x-1}\cdot 3 \;=\;3^{x-\frac{1}{2}}\cdot 2^2$

$\displaystyle \text{Divide by }2^2\!\cdot\!3\!:\;\;2^{2x-3} \;=\;3^{x-\frac{3}{2}}$ .[1]


$\displaystyle \text{Take logs: }\;\ln\left(2^{2x-3}\right) \;=\;\ln\left(3^{x-\frac{3}{2}}\right) \quad\Rightarrow\quad (2x-3)\!\cdot\log(2) \;=\;\left(x - \frac{3}{2}\right)\!\cdot\!\log(3)$

. . $\displaystyle 2x\cdot\ln(2) - 3\cdot\ln(2) \;=\;x\cdot\ln(3) - \frac{3}{2}\cdot\ln(3) \quad\Rightarrow\quad 2x\cdot\ln(2) - x\cdot\ln(3) \;=\;3\cdot\ln(2) - \frac{3}{2}\cdot\ln(3)$

$\displaystyle \text{Factor: }\;x\left[2\!\cdot\ln(2) - \ln(3)\right] \;=\;3\!\cdot\!\ln(2) - \frac{3}{2}\!\cdot\!\ln(3)$

$\displaystyle \text{Therefore: }\;x \;=\;\frac{3\ln(2) - \frac{3}{2}\ln(3)}{2\ln(2) - \ln(3)}$


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


$\displaystyle \text{On my calculator, the answer came out: }1.500000001$

. . . . $\displaystyle \text{and it turns out that }x = \frac{3}{2} \text{ is indeed the solution!}$

$\displaystyle \text{If the answer is }rational\text{, could we have found the answer without resorting to logs?}$


$\displaystyle \text{Well, there was one opportunity . . . back at }$ [1]

. . $\displaystyle \text{We had: }\;2^{2x-3} \:=\:3^{x-\frac{3}{2}}$


$\displaystyle \text{We could have }seen\text{ that the equation holds if both exponents were zero.}$

. . $\displaystyle \text{But }I\text{ didn't see it . . . did you?}$

 

[oded244]

Re: help with exponentials

nop.
Thanks for the solution.