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Mathematics 9 Online
OpenStudy (anonymous):

how do you solve 4^x + 2^(x+2)-5=0? I know the answer is supposed to be 0 but I dont know how you get that. Can you please show all steps?

OpenStudy (anonymous):

i think a good first step would be to write \[4^x+4\times 2^x-5=0\] because \(2^{x+2}=2^x\times 2^2=4\times 2^x\)

OpenStudy (anonymous):

a good next step is \[4^x=2^{2x}\] making your equation \[2^{2x}+4\times 2^x-5=0\] so you have a quadratic equation in \(2^x\) which you can probably solve by factoring

OpenStudy (anonymous):

\[(2^x-1)(2^x+5)=0\] \[2^x=1\] or\[2^x=-5\] second one is impossible, first one is true only if \(x=0\)

OpenStudy (anonymous):

thank you. Can most problems like this be solved this way?

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