solve 3^(2x^2+1) - 3^(x^2+2) = 3^(x^2) - 3 and
- Thread starter oded244
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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"?oded244 said:
Please reply with clarification, showing all of your work and reasoning. Thank you!
Eliz.
Re: help with exponentials
Hello, oded244!
\(\displaystyle \text{We have: }\;3^{2x^2+1} - 3^{x^2+2} - 3^{x^2} + 3 \;=\;0\)
\(\displaystyle \text{Factor: }\;3^{x^2+2}\left(3^{x^2-1} - 1\right) - 3\left(3^{x^2-1} - 1\right) \;=\;0\)
\(\displaystyle \text{Factor: }\;\left(3^{x^2+2} - 3\right)\,\left(3^{x^2-1} - 1\right) \;=\;0\)
\(\displaystyle \text{We have: }\;3^{x^2+2} - 3 \:=\:0 \quad\Rightarrow\quad 3^{x^2+2}\:=\:3^1\)
. . \(\displaystyle x^2+2 \:=\:1 \quad\Rightarrow\quad x^2\:=\:-1\quad\hdots\text{ no real roots}\)
\(\displaystyle \text{And we have: }\;3^{x^2-1}-1 \:=\:0 \quad\Rightarrow\quad 3^{x^2-1} \:=\:1 \quad\Rightarrow\quad 3^{x^2-1} \:=\:3^0\)
. . \(\displaystyle x^2-1 \:=\:0\quad\Rightarrow\quad x^2 \:=\:1 \quad\Rightarrow\quad\boxed{ x \:=\:\pm 1}\)
Hello, oded244!
\(\displaystyle \text{We have: }\;3^{2x^2+1} - 3^{x^2+2} - 3^{x^2} + 3 \;=\;0\)
\(\displaystyle \text{Factor: }\;3^{x^2+2}\left(3^{x^2-1} - 1\right) - 3\left(3^{x^2-1} - 1\right) \;=\;0\)
\(\displaystyle \text{Factor: }\;\left(3^{x^2+2} - 3\right)\,\left(3^{x^2-1} - 1\right) \;=\;0\)
\(\displaystyle \text{We have: }\;3^{x^2+2} - 3 \:=\:0 \quad\Rightarrow\quad 3^{x^2+2}\:=\:3^1\)
. . \(\displaystyle x^2+2 \:=\:1 \quad\Rightarrow\quad x^2\:=\:-1\quad\hdots\text{ no real roots}\)
\(\displaystyle \text{And we have: }\;3^{x^2-1}-1 \:=\:0 \quad\Rightarrow\quad 3^{x^2-1} \:=\:1 \quad\Rightarrow\quad 3^{x^2-1} \:=\:3^0\)
. . \(\displaystyle x^2-1 \:=\:0\quad\Rightarrow\quad x^2 \:=\:1 \quad\Rightarrow\quad\boxed{ x \:=\:\pm 1}\)
Re: help with exponentials
Hello again, oded244!
The second problem is a killer . . .
\(\displaystyle \text{We have: }\;\left(2^2\right)^x + 2^{2x-1} \;=\;3^{\frac{2x+1}{2}} + 3^{\frac{2x-1}{2}} \quad\Rightarrow\quad 2^{2x} + 2^{2x-1} \;=\;3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}\)
\(\displaystyle \text{Factor: }\;2^{2x-1}(2 + 1) \;=\;3^{x-\frac{1}{2}}(3 + 1) \quad\Rightarrow\quad 2^{2x-1}\cdot 3 \;=\;3^{x-\frac{1}{2}}\cdot 2^2\)
\(\displaystyle \text{Divide by }2^2\!\cdot\!3\!:\;\;2^{2x-3} \;=\;3^{x-\frac{3}{2}}\) .[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]
. . \(\displaystyle \text{We had: }\;2^{2x-3} \:=\:3^{x-\frac{3}{2}}\)
\(\displaystyle \text{We could have }seen\text{ that the equation holds if both exponents were zero.}\)
. . \(\displaystyle \text{But }I\text{ didn't see it . . . did you?}\)
Hello again, oded244!
The second problem is a killer . . .
\(\displaystyle \text{We have: }\;\left(2^2\right)^x + 2^{2x-1} \;=\;3^{\frac{2x+1}{2}} + 3^{\frac{2x-1}{2}} \quad\Rightarrow\quad 2^{2x} + 2^{2x-1} \;=\;3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}\)
\(\displaystyle \text{Factor: }\;2^{2x-1}(2 + 1) \;=\;3^{x-\frac{1}{2}}(3 + 1) \quad\Rightarrow\quad 2^{2x-1}\cdot 3 \;=\;3^{x-\frac{1}{2}}\cdot 2^2\)
\(\displaystyle \text{Divide by }2^2\!\cdot\!3\!:\;\;2^{2x-3} \;=\;3^{x-\frac{3}{2}}\) .[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]
. . \(\displaystyle \text{We had: }\;2^{2x-3} \:=\:3^{x-\frac{3}{2}}\)
\(\displaystyle \text{We could have }seen\text{ that the equation holds if both exponents were zero.}\)
. . \(\displaystyle \text{But }I\text{ didn't see it . . . did you?}\)