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Solve for x: e^2x + 4^e-2x = 4.
x =
I followed the steps sorban gave me for a similar problem and I got (1/2) In 2
x =
I followed the steps sorban gave me for a similar problem and I got (1/2) In 2
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- Feb 4, 2004
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Is your class really using "eye-enn" for the natural logarithm, instead of "ell-enn"? How odd.... :shock:
To check the solution to any "solving" exercise, plug it back into the original problem. Assuming, for instance, the first term to be "e^(2x) = e<sup>2x</sup>", rather than "e^2x = e<sup>2</sup>x", then:
. . . . .e<sup>2(1/2)ln(2)</sup> + 4<sup>e - 2(1/2)(ln(2))</sup> ?=? 4
. . . . .e<sup>ln(2)</sup> + 4<sup>e - ln(2)</sup> ?=? 4
. . . . .2 + 4<sup>e</sup> / 4<sup>ln(2)</sup> ?=? 4
. . . . .18.5673309777... ?=? 4
Since this doesn't check, either the ambiguity of your formatting has confused the original equation, or else your solution is incorrect. Sorry.
Eliz.
To check the solution to any "solving" exercise, plug it back into the original problem. Assuming, for instance, the first term to be "e^(2x) = e<sup>2x</sup>", rather than "e^2x = e<sup>2</sup>x", then:
. . . . .e<sup>2(1/2)ln(2)</sup> + 4<sup>e - 2(1/2)(ln(2))</sup> ?=? 4
. . . . .e<sup>ln(2)</sup> + 4<sup>e - ln(2)</sup> ?=? 4
. . . . .2 + 4<sup>e</sup> / 4<sup>ln(2)</sup> ?=? 4
. . . . .18.5673309777... ?=? 4
Since this doesn't check, either the ambiguity of your formatting has confused the original equation, or else your solution is incorrect. Sorry.
Eliz.
Hello, AmandaLynn04!
You typed the equation incorrectly . . . I think I know what it said.
Multiply by \(\displaystyle e^{2x}:\;\;e^{4x}\,+\,4\;=\;4e^{2x}\)
We have: \(\displaystyle \;e^{4x}\,-\,4e^{2x}\,+\,4\;=\;0\)
. . which factors: \(\displaystyle \;(e^{2x}\,-\,2)(e^{2x}\,-\,2)\;=\;0\)
. . and we have: \(\displaystyle \;e^{2x}\,-\,2\;=\;0\;\;\Rightarrow\;\;e^{2x}\,=\,2\;\;\Rightarrow\;\;2x\,=\,\ln 2\)
Therefore: \(\displaystyle \;x\:=\:\frac{1}{2}\ln2\)
You typed the equation incorrectly . . . I think I know what it said.
If I'm right about the equation, you're correct . . . nice work!
Multiply by \(\displaystyle e^{2x}:\;\;e^{4x}\,+\,4\;=\;4e^{2x}\)
We have: \(\displaystyle \;e^{4x}\,-\,4e^{2x}\,+\,4\;=\;0\)
. . which factors: \(\displaystyle \;(e^{2x}\,-\,2)(e^{2x}\,-\,2)\;=\;0\)
. . and we have: \(\displaystyle \;e^{2x}\,-\,2\;=\;0\;\;\Rightarrow\;\;e^{2x}\,=\,2\;\;\Rightarrow\;\;2x\,=\,\ln 2\)
Therefore: \(\displaystyle \;x\:=\:\frac{1}{2}\ln2\)