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Please help with (e^x)^x*e^28=e^11x
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\[(e ^{x})^{x}\times e ^{28}=e ^{11x}\] Is this it?
Yes
Take logs of both sides as follows:\[x ^{2}+28=11x\] Reaarange to form a quadratic and solve: \[x ^{2}-11x+28=0\] (x - 7)(x - 4) = 0 x = 7 and x = 4
Very smooth work there.....
Thx citizen :)
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I got ut right then, how bout 4^-x= 1/256, and 8^x-1.=64^2x
These are a little different. Note that the bases on both of these problems are compatible. I'll do the first one, you should be able to follow on the second.\[4^{-x}= 1/256 \implies (2^2)^{-x}=2^{-8} \implies -2x=-8 \implies x=4\]
kropot72, not right on this one.
Sorry. Please give medal to AnimalAin
No sweat. Just trying to be a team player here....
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vanessam: Have you got that?
?
-1/3
Nice job.
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