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2^(x^(3)-x)=1
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\[2^{x ^{3}-x}=1\]
how about the ones where an exponent has an exponent
solve the equation analytically.
solved it... how about \[3^{7x}=81^{4-2x}\]
\[7x \log 3= 81^{4-2x}\]
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can i write this like this
\[3^{4(4-2x)}\]
i get the same logs and i can just write 7x=4-2x
5x=4 x=4/5
i got it wrong @radar
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its supposed to be x=16/15
i gotttttt ittttt aaaayeeeee!!!!!!
Sorry I lead you astray, but glad you worked it out, did you use logs?
Oh!, never mind this looks like a different problem. I am studying it now.
|dw:1365101159216:dw|
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