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if c^2^x-2c^x=8 then x = ? can anyone help please
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gimmick is to write this as a quadratic equation in \[c^x\] since \[c^{2x}=(c^x)^2\]
i think i may have made a mistake all the c's replace them with the letter e
in other words solve \[z^2-2z-8=0\] \[(z-4)(z+2)=0\] \[z=4,z=-2\]
now replace oh ok
\[e^x=4, e^x=-2\] and \[e^x=-2\] is not possible since \[e^x>0\] so only solve \[e^x=4\] which in equivalent logarithmic form says \[x=\ln(4)\]
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oklay thanks youve been a great help :)
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