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OpenStudy (anonymous):

Had me baffled for a while 2^x + 2^(4-x) = 17

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OpenStudy (anonymous):

\[2^x+\frac{2^4}{2^x}=17\] then multiply by \[2^x\] and get a quadratic equation i think

OpenStudy (anonymous):

I am getting x=0,4

OpenStudy (anonymous):

Go on then...

OpenStudy (anonymous):

I know u can "see2 it's 0 and 4, what are the steps...

OpenStudy (anonymous):

\[2^x = a\]

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OpenStudy (anonymous):

and Now try solving quadratic way

OpenStudy (anonymous):

I eventually hit on2^x + 2^(x-4) = 2^4 + 2^0 but I'm not convinced...

OpenStudy (anonymous):

\[a^2 - 17a +16=0\]

OpenStudy (anonymous):

get \[2^{2x}+16=17\times 2^x\] or \[2^{2x}-17\times 2^x+16=0\] a quadratic in \[2^x\] solve \[z^2-17z+16=0\] get \[z=1, z = 16\] so \[2^x=1\iff x=0\] or \[2^x=16\iff x=4\]

OpenStudy (anonymous):

TV time, just thought it was amusing and no-one was answering it....

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